Ta có \(f(x)\) và \(g(x)\) là các hàm số liên tục tại \(x=1\). Do đó, hàm số \(2f(x)+g(x)\) cũng liên tục tại \(x=1\). Từ đó, ta có

\(\lim\limits_{x\to 1}{[2f(x)-g(x)]}=2f(1)-g(1)\Leftrightarrow 3=2.2-g(1)\Leftrightarrow g(1)=1.\)

Vậy \(g(1)=1.\)

}

+) \( \lim \limits_{x \to 1^+} f(x) =\lim \limits_{x \to 1^+} x=1\).

+) \( \lim \limits_{x \to 1^-} f(x) =\lim \limits_{x \to 1^-} (-x^2)=-1^2=-1\).

Vì \( \lim \limits_{x \to 1^+} f(x)=1\neq -1=\lim \limits_{x \to 1^-} f(x) \) nên không tồn tại giới hạn \( \lim \limits_{x \to 1} f(x) \).

}

a) Ta có \(\mathscr{D}_f=\mathbb{R}\setminus \{1\}\) và \(\mathscr{D}_g=\mathbb{R}\).

Do tập xác định của hai hàm số \(f(x)\) và \(g(x)\) khác nhau nên \(f(x)\ne g(x)\).

\underline{Cách khác:} Do \(f(x)\) không xác định, \(g(1)=2\) nên \(f(x)\ne g(x)\).

b) Ta có \(\lim \limits_{x \rightarrow 1} f(x)=\lim \limits_{x \rightarrow 1} \displaystyle\frac{(x-1)(x+1)}{x-1}=\lim \limits_{x \rightarrow 1} (x+1)=\lim \limits_{x \rightarrow 1} g(x)\).

}

Ta có

\begin{eqnarray*}\lim\limits_{x \to+\infty} \displaystyle\frac{2 x+1}{x-1}&=&\lim\limits_{x \to+\infty} \displaystyle\frac{x\left(2+\displaystyle\frac{1}{x}\right)}{x\left(1-\displaystyle\frac{1}{x}\right)}\\&=&\lim\limits_{x \to+\infty} \displaystyle\frac{2+\displaystyle\frac{1}{x}}{1-\displaystyle\frac{1}{x}}\\&=&\displaystyle\frac{\lim\limits_{x \to+\infty} 2+\lim\limits_{x \to+\infty} \displaystyle\frac{1}{x}}{\lim\limits_{x \to+\infty} 1-\lim\limits_{x \to+\infty} \displaystyle\frac{1}{x}}\\&=&\displaystyle\frac{2+0}{1-0}\\&=&2.\end{eqnarray*}

}

a) \(\underset{x \rightarrow-3}\lim \left(4 x^2-5 x+6\right)=4(-3)^2-5(-3)+6=57\).

b) \(\underset{x \rightarrow 2}\lim \displaystyle\frac{2 x^2-5 x+2}{x-2}=\underset{x \rightarrow 2}\lim \displaystyle\frac{(2 x-1)(x-2)}{x-2}=\underset{x \rightarrow 2}\lim (2 x-1)=3\).

c) \(\underset{x \rightarrow 4}\lim \displaystyle\frac{\sqrt{x}-2}{x^2-16}=\underset{x \rightarrow 4}\lim \displaystyle\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)}=\underset{x \rightarrow 4}\lim \displaystyle\frac{1}{(\sqrt{x}+2)(x+4)}=\displaystyle\frac{1}{32}\).

}

a) \(\underset{x \rightarrow-\infty}\lim \displaystyle\frac{6 x+8}{5 x-2}=\displaystyle\frac{6}{5}\).

b) \(\underset{x \rightarrow+\infty}\lim \displaystyle\frac{6 x+8}{5 x-2}=\displaystyle\frac{6}{5}\).

c) \(\underset{x \rightarrow-\infty}\lim \displaystyle\frac{\sqrt{9 x^2-x+1}}{3 x-2}=\underset{x \rightarrow-\infty}\lim \displaystyle\frac{-x \sqrt{9-\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x^2}}}{x\left(3-\displaystyle\frac{2}{x}\right)}=-1\).

d) \(\underset{x \rightarrow+\infty}\lim \displaystyle\frac{\sqrt{9 x^2-x+1}}{3 x-2}=\underset{x \rightarrow+\infty}\lim \displaystyle\frac{x \sqrt{9-\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x^2}}}{x\left(3-\displaystyle\frac{2}{x}\right)}=1\).

e) Vì \( \underset{x \rightarrow 2^{-}}\lim \left(3 x^2+4\right)=16>0 ;\underset{x \rightarrow 2^{-}}\lim (x+2)=0 \) và \(x \rightarrow-2^{-} \Rightarrow x+2 < 0\)

nên \\\(\underset{x \rightarrow-2^{-}}\lim \displaystyle\frac{3 x^2+4}{2 x+4}=-\infty\)

f) Vì \( \underset{x \rightarrow 2^{+}}\lim \left(3 x^2+4\right)=16>0 ;\underset{x \rightarrow 2^{+}}\lim (x+2)=0 \) và \(x \rightarrow-2^{+} \Rightarrow x+2>0\)

nên \\\(\underset{x \rightarrow-2^{-}}\lim \displaystyle\frac{3 x^2+4}{2 x+4}=+\infty\)

}

a) \(\lim \limits_{x \to-1}\left(3 x^2-x+2\right)=3(-1)^2-(-1)+2=6\).

b) \(\lim \limits_{x \to 4} \displaystyle\frac{x^2-16}{x-4}=\lim \limits_{x \to 4} (x+4)=8\).

c) \(\lim \limits_{x \to 2} \displaystyle\frac{3-\sqrt{x+7}}{x-2}=\lim \limits_{x \to 2} \displaystyle\frac{9-(x+7)}{(x-2)\left(3+\sqrt{x+7}\right)}=\lim \limits_{x \to 2} \displaystyle\frac{-1}{3+\sqrt{x+7}}=-\displaystyle\frac{1}{6}\).

}

a)

\begin{eqnarray*}&&\lim\limits_{x \to 2}\left(x^2-4 x+3\right)\\ &=&\lim\limits_{x \to 2} x^2-\lim\limits_{x \to 2} (4x)+\lim\limits_{x \to 2}3 =4-8+3=-1.\end{eqnarray*}

b)

\begin{eqnarray*}&&\lim\limits_{x \to 3} \displaystyle\frac{x^2-5 x+6}{x-3}\lim\limits_{x \to 3}\displaystyle\frac{(x-2)(x-3)}{x-3}=\lim\limits_{x \to 3}(x-2)\\ &=&\lim\limits_{x \to 3}x-\lim\limits_{x \to 3}2=3-2 =1.\end{eqnarray*}

c)

\begin{eqnarray*}&&\lim\limits_{x \to 1} \displaystyle\frac{\sqrt{x}-1}{x-1}=\lim\limits_{x \to 1}\displaystyle\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ &=&\lim\limits_{x \to 1}\displaystyle\frac{1}{\sqrt{x}+1}=\displaystyle\frac{\lim\limits_{x \to 1}1}{\lim\limits_{x \to 1}\sqrt{x}+\lim\limits_{x \to 1}1}\\ &=&\displaystyle\frac{1}{1+1}=\displaystyle\frac{1}{2}.\end{eqnarray*}

}

a)

\begin{eqnarray*}&&\lim\limits_{x \to+\infty} \displaystyle\frac{9 x+1}{3 x-4}=\lim\limits_{x \to+\infty}\displaystyle\frac{x\left(9+\displaystyle\frac{1}{x}\right)}{x\left(3-\displaystyle\frac{4}{x}\right)}\\ &=&\lim\limits_{x \to+\infty}\displaystyle\frac{9+\displaystyle\frac{1}{x}}{3-\displaystyle\frac{4}{x}}=\displaystyle\frac{\lim\limits_{x \to+\infty}9+\lim\limits_{x \to+\infty}\displaystyle\frac{1}{x}}{\lim\limits_{x \to+\infty}3-\lim\limits_{x \to+\infty}\displaystyle\frac{4}{x}}\\ &=&\displaystyle\frac{9+0}{3-0}=3.\end{eqnarray*}

b)

\begin{eqnarray*}&&\lim\limits_{x \to-\infty} \displaystyle\frac{7 x-11}{2 x+3}=\lim\limits_{x \to-\infty}\displaystyle\frac{x\left(7-\displaystyle\frac{11}{x}\right)}{x\left(2+\displaystyle\frac{3}{x}\right)}\\ &=&\lim\limits_{x \to-\infty}=\displaystyle\frac{7-\displaystyle\frac{11}{x}}{2+\displaystyle\frac{3}{x}}\\ &=&\displaystyle\frac{\lim\limits_{x \to-\infty}7-\lim\limits_{x \to-\infty}\displaystyle\frac{11}{x}}{\lim\limits_{x \to-\infty}2+\lim\limits_{x \to-\infty}\displaystyle\frac{3}{x}}=\displaystyle\frac{7-0}{2+0}=\displaystyle\frac{7}{2}.\end{eqnarray*}

c)

\begin{eqnarray*}&&\lim\limits_{x \to+\infty} \displaystyle\frac{\sqrt{x^2+1}}{x}=\lim\limits_{x \to+\infty}\displaystyle\frac{x\sqrt{1+\displaystyle\frac{1}{x^2}}}{x}\\ &=&\lim\limits_{x \to+\infty}\sqrt{1+\displaystyle\frac{1}{x^2}} =\sqrt{\lim\limits_{x \to+\infty}1+\lim\limits_{x \to+\infty}\displaystyle\frac{1}{x^2}}\\ &=&\sqrt{1+0}=1.\end{eqnarray*}

d)

\begin{eqnarray*}&&\lim\limits_{x \to-\infty} \displaystyle\frac{\sqrt{x^2+1}}{x}=\lim\limits_{x \to-\infty}\displaystyle\frac{-x\sqrt{1+\displaystyle\frac{1}{x^{2}}}}{x}\\ &=&\lim\limits_{x \to-\infty}-\sqrt{1+\displaystyle\frac{1}{x^{2}}}\\ &=&-\sqrt{\lim\limits_{x \to-\infty}1+\lim\limits_{x \to-\infty}\displaystyle\frac{1}{x^2}}=-\sqrt{1+0}=-1.\end{eqnarray*}

e) \(\lim\limits_{x \to 6^-} \displaystyle\frac{1}{x-6}=-\infty\).

f) \(\lim\limits_{x \to 7^+} \displaystyle\frac{1}{x-7}=+\infty\).

}

Ta có \(\lim\limits_{x \to 2^+} \displaystyle\frac{1}{x-2}=+\infty\).

}

Ta có \(\lim\limits_{x \to+\infty} x^3=+\infty; \lim\limits_{x \to-\infty} x^3=-\infty\).

}

Với dãy số \(\left(x_n\right)\) bất kì, \(x_n < 2\) và \(x_n \to 2\), ta có

\begin{eqnarray*}\lim\limits_{x_n \to 2^-} \sqrt{2-x_n}&=&\sqrt{\lim\limits_{x_n \to 2^-}\left(2-x_n\right)}\\&=&\sqrt{2-\lim\limits_{x_n \to 2^-} x_n}\\&=&\sqrt{2-2}\\&=&0.\end{eqnarray*}

Vậy \(\lim\limits_{x_n \to 2^-} \sqrt{2-x}=0\).

}

Ta có\(\lim\limits_{x \to 0^-} f(x)=-1\) và \(\lim\limits_{x \to 0^+} f(x)=1\).\\ Suy ra \(\lim\limits_{x \to 0^-} f(x) \neq \lim\limits_{x \to 0^+} f(x)\).

Vậy không tồn tại \(\lim\limits_{x \to 0} f(x)\).

}

a)

\begin{eqnarray*}\lim\limits_{x \to 2}\left(x^2+x-6\right)&=&\lim\limits_{x \to 2} x^2+\lim\limits_{x \to 2} x-\lim\limits_{x \to 2} 6\\&=&4+2-6\\&=&0.\end{eqnarray*}

b)

\begin{eqnarray*}\lim\limits_{x \to 1} \displaystyle\frac{x^2+2 x+3}{2 x-1}&=&\displaystyle\frac{\lim\limits_{x \to 1}\left(x^2+2 x+3\right)}{\lim\limits_{x \to 1}(2 x-1)}\\&=&\displaystyle\frac{\lim\limits_{x \to 1} x^2+\lim\limits_{x \to 1}(2 x)+\lim\limits_{x \to 1} 3}{\lim\limits_{x \to 1}(2 x)-\lim\limits_{x \to 1} 1}\\&=&\displaystyle\frac{1+2+3}{2-1}\\&=&6.\end{eqnarray*}

}

a) \(\lim \limits_{x \to 4^{+}} \displaystyle\frac{1}{x-4}=+\infty\).

do \(\begin{cases}\lim \limits_{x \to 4^{+}} 1=1>0\\ \lim \limits_{x \to 4^{+}} (x-4)=0\\ \text{Khi } x\to 4^+\Leftrightarrow x>4\Leftrightarrow x-4>0.\end{cases}\)

b) \(\lim \limits_{x \to 2^{+}} \displaystyle\frac{x}{2-x} =-\infty\).

do \(\begin{cases}\lim \limits_{x \to 2^{+}} x=2>0\\ \lim \limits_{x \to 2^{+}} (2-x)=0\\ \text{Khi } x\to 2^+\Leftrightarrow x>2\Leftrightarrow 2-x < 0.\end{cases}\)

}

a) \(\lim \limits_{x \to+\infty} \displaystyle\frac{-x+2}{x+1}=\lim \limits_{x \to+\infty} \displaystyle\frac{-1+\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}=-1\).

b) \(\lim \limits_{x \to-\infty} \displaystyle\frac{x-2}{x^2}=\lim \limits_{x \to-\infty} \displaystyle\frac{\displaystyle\frac{1}{x}-\displaystyle\frac{2}{x^2}}{1}=0\).

}

a) \( \lim \limits_{x \to -1^+} \displaystyle\frac{1}{x+1} \);

Ta có \( \lim \limits_{x \to -1^+} \displaystyle\frac{1}{x+1}=\lim \limits_{x \to -1^+} \displaystyle\frac{1}{x-(-1)}=+\infty \).

Vậy \( \lim \limits_{x \to -1^+} \displaystyle\frac{1}{x+1} =+\infty \).

b) \( \lim \limits_{x \to -\infty} (1-x^2)\);

Ta có \( (1-x^2)=x^2\cdot \left(\displaystyle\frac{1}{x^2}-1\right)\) và \( \lim \limits_{x \to -\infty} x^2=+\infty \); \quad \( \lim \limits_{x \to -\infty} \left(\displaystyle\frac{1}{x^2}-1\right)=-1 \).

Vậy \( \lim \limits_{x \to -\infty} (1-x^2)=-\infty\).

c) \( \lim \limits_{x \to 3^-} \displaystyle\frac{x}{3-x} \).

Ta có \( \displaystyle\frac{x}{3-x}=x\cdot \displaystyle\frac{-1}{x-3} \) và \( \lim \limits_{x \to 3^-} x =3\); \quad \( \lim \limits_{x \to 3^-} \displaystyle\frac{-1}{x-3}=+\infty \).

Vậy \( \lim \limits_{x \to 3^-} \displaystyle\frac{x}{3-x}=+\infty \).

}

a) \( \lim \limits_{x \to +\infty} \displaystyle\frac{4x+3}{2x}=\lim \limits_{x \to +\infty} \displaystyle\frac{4+\displaystyle\frac{3}{x}}{2}=\displaystyle\frac{4+0}{2}=2 \);

b) \( \lim \limits_{x \to -\infty} \displaystyle\frac{2}{3x+1}=\lim \limits_{x \to -\infty} \displaystyle\frac{\displaystyle\frac{2}{x}}{3+\displaystyle\frac{1}{x}}=\displaystyle\frac{0}{3+0}=0 \);

c) \( \lim \limits_{x \to +\infty} \displaystyle\frac{\sqrt{x^2+1}}{x+1}=\lim \limits_{x \to +\infty} \displaystyle\frac{\sqrt{x^2\left(1+\displaystyle\frac{1}{x^2}\right)}}{x\left(1+\displaystyle\frac{1}{x}\right)}=\lim \limits_{x \to +\infty} \displaystyle\frac{|x|\sqrt{\left(1+\displaystyle\frac{1}{x^2}\right)}}{x\left(1+\displaystyle\frac{1}{x}\right)}=\lim \limits_{x \to +\infty} \displaystyle\frac{\sqrt{1+\displaystyle\frac{1}{x^2}}}{1+\displaystyle\frac{1}{x}} =1\).

}

a) \( \lim \limits_{x \to -2} (x^2-7x+4)=\lim \limits_{x \to -2} x^2-7\lim \limits_{x \to -2} x +4=(-2)^2-7\cdot (-2)+4=22\);

b) \( \lim \limits_{x \to 3} \displaystyle\frac{x-3}{x^2-9} =\lim \limits_{x \to 3} \displaystyle\frac{1}{x+3}=\displaystyle\frac{1}{3+3}=\displaystyle\frac{1}{6}\);

c) \( \lim \limits_{x \to 1} \displaystyle\frac{3-\sqrt{x+8}}{x-1}=\lim \limits_{x \to 1} \displaystyle\frac{1-x}{(x-1)\left(3+\sqrt{x+8}\right)}=\lim \limits_{x \to 1} \displaystyle\frac{-1}{\left(3+\sqrt{x+8}\right)} =\displaystyle\frac{-1}{3+\sqrt{1+8}}=\displaystyle\frac{-1}{6}\).

}

a) \( \lim \limits_{x \to 3^-} \displaystyle\frac{2x}{x-3} \);

Ta có \( \lim \limits_{x \to 3^-} (2x)=2\lim \limits_{x \to 3^-}x=2\cdot (-3)=-6 \); \( \lim \limits_{x \to 3^-} \displaystyle\frac{1}{x-3}=-\infty \).

Do đó \( \lim \limits_{x \to 3^-} \displaystyle\frac{2x}{x-3}=\lim \limits_{x \to 3^-} \left[(2x)\cdot \displaystyle\frac{1}{x-3}\right]=+\infty \).

c) \( \lim \limits_{x \to +\infty} (3x-1) \).

Viết \( 3x-1 =x\left(3-\displaystyle\frac{1}{x}\right)\).

Ta có \( \lim \limits_{x \to +\infty} x=+\infty \); \( \lim \limits_{x \to +\infty} \left(3-\displaystyle\frac{1}{x}\right)=3-\lim \limits_{x \to +\infty} \displaystyle\frac{1}{x}=3-0=3 \).

Do đó \( \lim\limits_{x \to +\infty} (3x-1)=\lim \limits_{x \to +\infty} \left[x\left(3-\displaystyle\frac{1}{x}\right)\right]=+\infty\).

}

a) \( \lim \limits_{x \to +\infty} \displaystyle\frac{1-3x^2}{x^2+2x}=\lim\limits_{x \to +\infty}\displaystyle\frac{\displaystyle\frac{1}{x^2}-3}{1+\displaystyle\frac{2}{x}}=\displaystyle\frac{\lim \limits_{x \to +\infty}\left(\displaystyle\frac{1}{x^2}-3\right)}{\lim \limits_{x \to +\infty}\left(1+\displaystyle\frac{2}{x}\right)}=\displaystyle\frac{\lim \limits_{x \to +\infty}\displaystyle\frac{1}{x^2}-\lim \limits_{x \to +\infty} 3}{\lim \limits_{x \to +\infty} 1+\lim \limits_{x \to +\infty} \displaystyle\frac{2}{x}}=\displaystyle\frac{0-3}{1+0}=-3\);

b) \( \lim \limits_{x \to -\infty} \displaystyle\frac{2}{x+1}=\lim \limits_{x \to -\infty}\displaystyle\frac{\displaystyle\frac{2}{x}}{1+\displaystyle\frac{1}{x}}=\displaystyle\frac{\lim \limits_{x \to -\infty}\displaystyle\frac{2}{x}}{\lim \limits_{x \to -\infty}\left(1+\displaystyle\frac{1}{x}\right)} =\displaystyle\frac{2\cdot \lim \limits_{x \to -\infty}\displaystyle\frac{1}{x}}{\lim \limits_{x \to -\infty} 1+\lim \limits_{x \to -\infty} \displaystyle\frac{1}{x}}=\displaystyle\frac{0}{1+0}=0\).

}

a) Đặt \( g(x)=2x^2-x \).

Hàm số \( y=g(x) \) xác định trên \( \mathbb{R}\).

Giả sử \( (x_n) \) là dãy số bất kì, thỏa mãn \( x_n\to 3 \) khi \( n\to +\infty \). Ta có

\(\lim g(x_n)=\lim (2x_n^2-x_n)=2\cdot \left(\lim x_n\right)^2-\lim x_n=2\cdot 3^2-3=15.\)

Vậy \( \lim \limits_{x \to 3} g(x)=15 \).

b) Đặt \( h(x)=\displaystyle\frac{x^2+2x+1}{x+1} \).

Hàm số \( y=h(x) \) xác định trên \( \mathbb{R}\setminus\{-1\}\).

Giả sử \( (x_n) \) là dãy số bất kì, thỏa mãn \( x_n\neq -1 \) với mọi \( n\) và \( x_n\to -1 \) khi \( n\to +\infty \). Ta có

\(\lim h(x_n)=\lim \displaystyle\frac{x_n^2+2x_n+1}{x_n+1}=\lim (x_n+1)=\lim x_n+1=-1+1=0.\)

Vậy \( \lim \limits_{x \to -1} h(x)=0 \).

}

a) \( \lim \limits_{x \to 1} (x^2-4x+2)=\lim \limits_{x \to 1} x^2-\lim \limits_{x \to 1} (4x)+\lim \limits_{x \to 1} 2=1^2-4\lim \limits_{x \to 1} x+2=1-4\cdot 1+2=-1\);

b) \( \lim \limits_{x \to 2} \displaystyle\frac{3x-2}{2x+1}=\displaystyle\frac{\lim \limits_{x \to 2} (3x-2)}{\lim \limits_{x \to 2} (2x+1)}=\displaystyle\frac{3\lim \limits_{x \to 2} x-2}{2\lim \limits_{x \to 2} x+1}=\displaystyle\frac{3 \cdot 2-2}{2\cdot 2+1}=\displaystyle\frac{4}{5}\).

}

a) \( \lim \limits_{x \to 2} \displaystyle\frac{x^2-4}{x-2}=\lim \limits_{x \to 2} \displaystyle\frac{(x-2)(x+2)}{x-2}=\lim \limits_{x \to 2} (x+2)=\lim \limits_{x \to 2} x+\lim \limits_{x \to 2} 2=2+2=4\).

b) \( \lim \limits_{x \to 3} \displaystyle\frac{\sqrt{x+1}-2}{x-3}\).

\begin{eqnarray*}\lim \limits_{x \to 3} \displaystyle\frac{\sqrt{x+1}-2}{x-3}&=&\lim \limits_{x \to 3} \displaystyle\frac{\left(\sqrt{x+1}-2\right)\left(\sqrt{x+1}+2\right)}{(x-3)\left(\sqrt{x+1}+2\right)}\quad \left(\text{nhân cả tử và mẫu cho }\left(\sqrt{x+1}+2\right)\right)\\&=&\lim \limits_{x \to 3} \displaystyle\frac{(x+1)-4}{(x-3)\left(\sqrt{x+1}+2\right)}=\lim \limits_{x \to 3} \displaystyle\frac{1}{\sqrt{x+1}+2}\\&=& \displaystyle\frac{1}{\lim \limits_{x \to 3} \left(\sqrt{x+1}+2\right)}=\displaystyle\frac{1}{\lim \limits_{x \to 3} \sqrt{x+}+2}\\&=& \displaystyle\frac{1}{\sqrt{\lim \limits_{x \to 3} (x+1)}+2}=\displaystyle\frac{1}{\sqrt{3+1}+2}=\displaystyle\frac{1}{4}.\end{eqnarray*}

}

a) \( \lim \limits_{x \to -2} (x^2+5x-2)=(\lim \limits_{x \to -2} x)^2+\lim \limits_{x \to -2} (5x)-\lim \limits_{x \to -2} 2=(-2)^2+5\cdot (-2)-2=-8\).

b) \( \lim \limits_{x \to 1} \displaystyle\frac{x^2-1}{x-1}=\lim \limits_{x \to 1} \displaystyle\frac{(x-1)(x+1)}{(x-1)}=\lim \limits_{x \to 1} (x+1)=\lim \limits_{x \to 1} x+1=1+1=2\).

}

\( \lim\limits_{x \to -\infty} \displaystyle\frac{x^2-3x}{2x^2+1}=\lim\limits_{x \to -\infty}\displaystyle\frac{1-\displaystyle\frac{3}{x}}{2+\displaystyle\frac{1}{x^2}}=\displaystyle\frac{\lim\limits_{x \to -\infty}\left(1-\displaystyle\frac{3}{x}\right)}{\lim\limits_{x \to -\infty}\left(2+\displaystyle\frac{1}{x^2}\right)}=\displaystyle\frac{1-\lim\limits_{x \to -\infty}\displaystyle\frac{3}{x}}{2+\lim \limits_{x \to -\infty}\displaystyle\frac{1}{x^2}}=\displaystyle\frac{1-0}{2+0}=\displaystyle\frac{1}{2}\).

}

a) \( \lim \limits_{x \to 2^+} \displaystyle\frac{1-2x}{x-2} \);

Ta có \( \lim \limits_{x \to 2^+} (1-2x)=1-2\lim \limits_{x \to 2^+}x=1-2\cdot 2=-3 \); \( \lim \limits_{x \to 2^+} \displaystyle\frac{1}{x-2}=+\infty \).

Do đó \( \lim \limits_{x \to 2^+} \displaystyle\frac{1-2x}{x-2}=\lim \limits_{x \to 2^+} \left[(1-2x)\cdot \displaystyle\frac{1}{x-2}\right]=-\infty \).

b) \( \lim \limits_{x \to -\infty} (x^2+1) \).

Viết \( x^2+1 =x^2\left(1+\displaystyle\frac{1}{x^2}\right)\).

Ta có \( \lim \limits_{x \to -\infty} x^2=+\infty \); \( \lim \limits_{x \to -\infty} \left(1+\displaystyle\frac{1}{x^2}\right)=1+\lim \limits_{x \to -\infty} \displaystyle\frac{1}{x^2}=1+0=1 \).

Do đó \( \lim\limits_{x \to -\infty} (x^2+1)=\lim \limits_{x \to -\infty} \left[x^2\left(1+\displaystyle\frac{1}{x^2}\right)\right]=+\infty\).

}

+) Giả sử \( (x_n) \) là dãy số bất kì, \( x_n < -1 \) và \( x_n \to -1\). Khi đó \( \lim f(x_n)=\lim (1-2x_n)=3 \).

Vậy \( \lim \limits_{x \to -1^-} f(x)=3\).

Giả sử \( (x_n) \) là dãy số bất kì, \( x_n>-1 \) và \( x_n \to -1\). Khi đó \( \lim f(x_n)=\lim (x_n^2+2)=3 \).

Vậy \( \lim \limits_{x \to -1^+} f(x)=3\).

+) Vì \( \lim \limits_{x \to -1^+} f(x) = \lim \limits_{x \to -1^-} f(x) \) nên tồn tại \( \lim \limits_{x \to -1} f(x) \) và \( \lim \limits_{x \to -1} f(x)=3 \).

}

a) Ta có

\begin{eqnarray*}\mathop {\lim }\limits_{x\to 7} \displaystyle\frac{\sqrt{x+2}-3}{x-7}&=&\mathop {\lim }\limits_{x\to 7} \displaystyle\frac{\left(\sqrt{x+2}\right)^2-3^2}{(x-7)\left(\sqrt{x+2}+3\right)}=\mathop {\lim }\limits_{x\to 7} \displaystyle\frac{x-7}{(x-7)\left(\sqrt{x+2}+3\right)}\\&=&\mathop {\lim }\limits_{x\to 7} \displaystyle\frac{1}{\sqrt{x+2}+3}=\displaystyle\frac{1}{6}.\end{eqnarray*}

b) Ta có

\begin{eqnarray*}\mathop {\lim }\limits_{x\to 1} \displaystyle\frac{x^{3}-1}{x^{2}-1}&=&\mathop {\lim }\limits_{x\to 1} \displaystyle\frac{(x-1)(x^2+x+1)}{(x-1)(x+1)}\\&=&\mathop {\lim }\limits_{x\to 1} \displaystyle\frac{x^2+x+1}{x+1}=\displaystyle\frac{3}{2}.\end{eqnarray*}

c) Ta có \(\mathop {\lim }\limits_{x\to 1} (2-x)=1;\)

\(\mathop {\lim }\limits_{x\to 1} (1-x)^2=0;\) \((1-x)^2 > 0,\) \(\forall x \neq 0\) nên \(\mathop {\lim }\limits_{x\to 1} \displaystyle\frac{2-x}{(1-x)^{2}}=+\infty.\)

d) Ta có

\begin{eqnarray*}\mathop {\lim }\limits_{x\to -\infty} \displaystyle\frac{x+2}{\sqrt{4 x^{2}+1}}&=&\mathop {\lim }\limits_{x\to -\infty} \displaystyle\frac{x+2}{|x|\sqrt{4 +\displaystyle\frac{1}{x^2}}}=\mathop {\lim }\limits_{x\to -\infty} \displaystyle\frac{x\left( 1+\displaystyle\frac{2}{x}\right)}{-x\sqrt{4 +\displaystyle\frac{1}{x^2}}}\\&=&\mathop {\lim }\limits_{x\to -\infty} \displaystyle\frac{ 1+\displaystyle\frac{2}{x}}{-\sqrt{4 +\displaystyle\frac{1}{x^2}}}=-\displaystyle\frac{1}{2}.\end{eqnarray*}

}

a) Ta có

\begin{eqnarray*}\lim \limits_{x \rightarrow+\infty} \displaystyle\frac{1-2 x}{\sqrt{x^{2}+1}}&=& -\lim \limits_{x \rightarrow+\infty}\sqrt{\displaystyle\frac{4x^2-4x+1}{x^2+1}}\\&=& -\lim \limits_{x \rightarrow+\infty}\sqrt{4-\displaystyle\frac{4x}{x^2+1}+\displaystyle\frac{1}{x^2+1}}\\&=& -\sqrt{4-\lim \limits_{x \rightarrow+\infty} \displaystyle\frac{4x}{x^2+1}+ \lim \limits_{x \rightarrow+\infty}\displaystyle\frac{1}{x^2+1}}\\&=& -\sqrt{4-\lim \limits_{x \rightarrow+\infty} \displaystyle\frac{\displaystyle\frac{4}{x}}{1+\displaystyle\frac{1}{x^2}}+ \lim \limits_{x \rightarrow+\infty}\displaystyle\frac{\displaystyle\frac{1}{x^2}}{1+\displaystyle\frac{1}{x^2}}}\\&=& -2.\end{eqnarray*}

b) Ta có

\begin{eqnarray*}\sqrt{x^{2}+x+2}-x&=& \displaystyle\frac{\left(\sqrt{x^{2}+x+2} \right)^2-x^2 }{\sqrt{x^{2}+x+2}+x}=\displaystyle\frac{x+2 }{\sqrt{x^{2}+x+2}+x}\\&=& \displaystyle\frac{x\cdot\left(1+\displaystyle\frac{2}{x} \right) }{x\cdot \left(\sqrt{1+\displaystyle\frac{1}{x}+\displaystyle\frac{2}{x^2}}+1 \right) }=\displaystyle\frac{1+\displaystyle\frac{2}{x}}{\sqrt{1+\displaystyle\frac{1}{x}+\displaystyle\frac{2}{x^2}}+1 }.\end{eqnarray*}

Khi đó

\(\lim \limits_{x \rightarrow+\infty}\left(\sqrt{x^{2}+x+2}-x\right)=\lim \limits_{x \rightarrow+\infty} \displaystyle\frac{1+\displaystyle\frac{2}{x}}{\sqrt{1+\displaystyle\frac{1}{x}+\displaystyle\frac{2}{x^2}}+1 }=\displaystyle\frac{1}{1+1}=\displaystyle\frac{1}{2}\).

}

Viết \(\displaystyle\frac{2}{(x-1)(x-2)}=\displaystyle\frac{2}{x-1} \cdot \displaystyle\frac{1}{x-2}\), ta có \(\lim \limits_{x \rightarrow 2^{+}} \displaystyle\frac{2}{x-1}=2>0\).

Hơn nữa \(\lim \limits_{x \rightarrow 2^{+}} \displaystyle\frac{1}{x-2}=+\infty\) do \(x-2>0\) khi \(x>2\).

Áp dụng quy tắc tìm giới hạn của tích, ta được \(\lim \limits_{x \rightarrow 2^{+}} \displaystyle\frac{2}{(x-1)(x-2)}=+\infty\).

Lí luận tương tự, ta có \(\lim \limits_{x \rightarrow 2^-} \displaystyle\frac{1}{x(1-x)}=-\infty\).}

a) \(\lim \limits_{x \rightarrow 0} \displaystyle\frac{(x+2)^{2}-4}{x}=\lim \limits_{x \rightarrow 0} \displaystyle\frac{x^2+4x}{x}=\lim \limits_{x \rightarrow 0} (x+4)=4\).

b) \(\lim \limits_{x \rightarrow 0} \displaystyle\frac{\sqrt{x^{2}+9}-3}{x^{2}}=\lim \limits_{x \rightarrow 0} \displaystyle\frac{x^2}{x^{2}\cdot \left(\sqrt{x^{2}+9}+3 \right) }=\lim \limits_{x \rightarrow 0} \displaystyle\frac{1}{\sqrt{x^{2}+9}+3}=\displaystyle\frac{1}{6}\).

}

Ta sử dụng quy tắc tìm giới hạn của thương. Rõ ràng, giới hạn của tử số \(\lim \limits_{x \rightarrow 0}(x+1)=1\).

Ngoài ra, mẫu số nhận giá trị dương với mọi \(x \neq 0\) và \(\lim \limits_{x \rightarrow 0} x^{2}=0\).

Do vậy \(\lim \limits_{x \rightarrow 0} \displaystyle\frac{x+1}{x^{2}}=+\infty\).

}

Ta có

\begin{eqnarray*}\lim \limits_{x \rightarrow-\infty} \displaystyle\frac{\sqrt{x^{2}+1}}{x}&=& \lim \limits_{x \rightarrow-\infty}\left(-\sqrt{\displaystyle\frac{x^{2}+1}{x^{2}}}\right)\\&=& -\lim \limits_{x \rightarrow-\infty} \sqrt{1+\displaystyle\frac{1}{x^{2}}}\\&=& -\sqrt{\lim \limits_{x \rightarrow-\infty}\left(1+\displaystyle\frac{1}{x^{2}}\right)}\\&=& -\sqrt{1+\lim \limits_{x \rightarrow-\infty} \displaystyle\frac{1}{x^{2}}}=-1.\end{eqnarray*}

}

Xét hàm số \(f(x)=\displaystyle\frac{1}{|x-1|}\). Lấy dãy số \(\left(x_{n}\right)\) bất kì sao cho \(x_{n} \neq 1\), \(x_{n} \rightarrow 1\).

Khi đó, \(\left|x_{n}-1\right| \rightarrow 0\).

Do đó \(f\left(x_{n}\right)=\displaystyle\frac{1}{\left|x_{n}-1\right|} \rightarrow+\infty\). Vậy \(\lim \limits_{x \rightarrow 1} \displaystyle\frac{1}{|x-1|}=+\infty\).

}

Viết \(\displaystyle\frac{1}{x(1-x)}=\displaystyle\frac{1}{x} \cdot \displaystyle\frac{1}{1-x}\), ta có \(\lim \limits_{x \rightarrow 1^{+}} \displaystyle\frac{1}{x}=1>0\).

Hơn nữa \(\lim \limits_{x \rightarrow 1^{+}} \displaystyle\frac{1}{1-x}=-\infty\) do \(1-x < 0\) khi \(x>1\).

Áp dụng quy tắc tìm giới hạn của tích, ta được \(\lim \limits_{x \rightarrow 1^{+}} \displaystyle\frac{1}{x(1-x)}=-\infty\).

Lí luận tương tự, ta có \(\lim \limits_{x \rightarrow 1^-} \displaystyle\frac{1}{x(1-x)}=+\infty\).

}

a) Ta có \(\lim \limits_{x \rightarrow 1^{+}} (x-2)=-1 < 0\).

Hơn nữa \(\lim \limits_{x \rightarrow 1^{+}} (x-1)=0\), và \(x-1>0\) khi \(x>1\).

Áp dụng quy tắc tìm giới hạn của thương, ta được \(\lim \limits_{x \rightarrow 1^{+}} \displaystyle\frac{x-2}{x-1}=-\infty\).

b) Ta có \(\lim \limits_{x \rightarrow 4^{-}} (x^{2}-x+1)=13>0\).

Hơn nữa \(\lim \limits_{x \rightarrow 4^{-}} (4-x)=0\), và \(4-x>0\) khi \(x < 4\).

Áp dụng quy tắc tìm giới hạn của thương, ta được \(\lim \limits_{x \rightarrow 4^{-}} \displaystyle\frac{x^{2}-x+1}{4-x}=+\infty\).

}

a) Ta có \(x\to 3^+ \Rightarrow x>3 \Rightarrow x-3>0.\) Vậy

\begin{eqnarray*}\mathop {\lim }\limits_{x\to 3^+} \displaystyle\frac{x^{2}-9}{|x-3|}&=&\mathop {\lim }\limits_{x\to 3^+} \displaystyle\frac{(x-3)(x+3)}{x-3}\\&=&\mathop {\lim }\limits_{x\to 3^+} (x+3)=6.\end{eqnarray*}

b) Ta có \(\mathop {\lim }\limits_{x\to 1^-} x=1;\) \(\mathop {\lim }\limits_{x\to 1^-} \sqrt{1-x}=0\) và \(\sqrt{1-x}>0\), \(\forall x < 1\) nên \(\mathop {\lim }\limits_{x\to 1^-} \displaystyle\frac{x}{\sqrt{1-x}}=+\infty.\)

}

a) Giả sử \( (x_n) \) là dãy số bất kì, \( x_n>0 \) và \( x_n \to 0\). Khi đó \( f(x_n)=1 \) nên \( \lim f(x_n)=\lim 1=1 \).

Vậy \( \lim \limits_{x \to 0^+} f(x)=1\).

Giả sử \( (x_n) \) là dãy số bất kì, \( x_n < 0 \) và \( x_n \to 0\). Khi đó \( f(x_n)=0 \) nên \( \lim f(x_n)=\lim 0= 0\).

Vậy \( \lim \limits_{x \to 0^-} f(x)=0\).

b) Vì \( \lim \limits_{x \to 0^+} f(x) \neq \lim \limits_{x \to 0^-} f(x) \) nên không tồn tại \( \lim \limits_{x \to 0} f(x) \).

}

+) \(\mathop {\lim }\limits_{x\to 0^-} \displaystyle\frac{|x|}{x}=\mathop {\lim }\limits_{x\to 0^-} \displaystyle\frac{-x}{x}=\mathop {\lim }\limits_{x\to 0^-} (-1)=-1.\)

+) \(\mathop {\lim }\limits_{x\to 0^+} \displaystyle\frac{|x|}{x}=\mathop {\lim }\limits_{x\to 0^+} \displaystyle\frac{x}{x}=\mathop {\lim }\limits_{x\to 0^+} 1=1.\)

Vậy \(\mathop {\lim }\limits_{x\to 0^-} \displaystyle\frac{|x|}{x}\ne \mathop {\lim }\limits_{x\to 0^+} \displaystyle\frac{|x|}{x}\) nên giới hạn \(\mathop {\lim }\limits_{x\to 0} \displaystyle\frac{|x|}{x}\) không tồn tại.

}

Ta có \(\lim \limits_{x \rightarrow 2^{+}} g(x)=\lim \limits_{x \rightarrow 2^{+}}\displaystyle\frac{(x-2)(x-3)}{x-2}\,\,(\text{do } x>2)=\lim \limits_{x \rightarrow 2^{+}} (x-3)=-1\).

Tương tự \(\lim \limits_{x \rightarrow 2^{-}} g(x)=\lim \limits_{x \rightarrow 2^{-}}-\displaystyle\frac{(x-2)(x-3)}{x-2}\,\,(\text{do } x < 2)=-\lim \limits_{x \rightarrow 2^{-}} (x-3)=1\).

}

a) Ta có \(\displaystyle\frac{1}{d}+\displaystyle\frac{1}{d^{\prime}}=\displaystyle\frac{1}{f} \Leftrightarrow d^{\prime}=\displaystyle\frac{d f}{d-f}.\)

Vậy \(\varphi(d)=\displaystyle\frac{d f}{d-f}\).

b) Vì \(\underset{d \rightarrow f^{+}}\lim df=f^2; \underset{d \rightarrow f^{+}}\lim (d-f)=0 ; d \rightarrow f^{+} \Rightarrow d-f>0 \) nên \( \underset{d \rightarrow f^{+}}\lim \displaystyle\frac{d f}{d-f}=+\infty\).

Vậy \(\underset{d \rightarrow f^{+}}\lim \varphi(d)=\underset{d \rightarrow f^{+}}\lim \displaystyle\frac{d f}{d-f}=+\infty\).

Ý nghĩa: Khi đặt vật nằm ngoài tiêu cự và tiến dần đến tiêu điểm thì cho ảnh thật ngược chiều với vật ở vô cùng.

Vì \(\underset{d \rightarrow f^{-}}\lim df=f^2; \underset{d \rightarrow f^{+}}\lim (d-f)=0 ; d \rightarrow f^{-} \Rightarrow d-f < 0 \) nên \( \underset{d \rightarrow f^{+}}\lim \displaystyle\frac{d f}{d-f}=-\infty\).

Vậy \(\underset{d \rightarrow f^{+}}\lim \varphi(d)=\underset{d \rightarrow f^{+}}\lim \displaystyle\frac{d f}{d-f}=-\infty\).

Ý nghĩa: Khi đặt vật nằm trong tiêu cự và tiến dần đến tiêu điểm thì cho ảnh ảo cùng chiều với vật và nằm ở vô cùng.

Vì không tồn tại \(\underset{d \rightarrow f^{+}}\lim \varphi(d)\) và \(\underset{d \rightarrow f^{-}}\lim \varphi(d)\) nên không tồn tại \(\underset{d \rightarrow f}\lim \varphi(d)\).

}

Ta có \(\lim \limits_{t \rightarrow 0^{+}} H(t)=1\) và \(\lim \limits_{t \rightarrow 0^{-}} H(t)=0\).

}

a) Sau thời gian \( t \) phút, số m\(^3\) nước trong hồ là \(200+2t \) (m\(^3\)).

Số kilôgam muối là \( 200\cdot 10=2000 \) (kg).

Nồng độ muối của nước trong hồ sau \( t \) phút khi bắt đầu bơm là

\(C(t)=\displaystyle\frac{2000}{200+2t}=\displaystyle\frac{1000}{100+t}\); \((\mathrm{kg/m}^3).\)

b) Khi \( t\to +\infty \), ta xét giới hạn

\(\lim \limits_{t \to +\infty} C(t)=\lim \limits_{t \to +\infty} \displaystyle\frac{1000}{100+t}=0.\)

}

Ta có

+) Với \(r < R,\) \(F(r)=\displaystyle\frac{G Mr}{R^{3}}\) là hàm liên tục.

+) Với \(r>R,\) \(F(r)=\displaystyle\frac{G M}{R^{2}}\) là hàm liên tục.

Tại \(r=R.\)

+) \(F(R)=\displaystyle\frac{G M}{R^{2}}. \)

+) \(\mathop {\lim }\limits_{r\to R^+}F(r)=\mathop {\lim }\limits_{r\to R^+}\displaystyle\frac{G M}{r^{2}}=\displaystyle\frac{G M}{R^{2}}.\)

+) \(\mathop {\lim }\limits_{r\to R^-}F(r)=\mathop {\lim }\limits_{r\to R^-}\displaystyle\frac{G M r}{R^{3}}=\displaystyle\frac{G M R}{R^{3}}=\displaystyle\frac{G M}{R^{2}}.\)

Ta có \(F(R)=\mathop {\lim }\limits_{r\to R^+}F(r)=\mathop {\lim }\limits_{r\to R^-}F(r)\) nên hàm số liên tục tại \(r=R\).

Vậy \(F(r)\) liên tục trên \(\mathbb{R}\).

}

a) \( \lim \limits_{d \to f^+} g(d) =\lim \limits_{d \to f^+} \displaystyle\frac{f}{1-\displaystyle\frac{f}{d}}\); \quad \( \lim \limits_{d \to f^+} f=f \); \quad \(\lim \limits_{d \to f^+}\displaystyle\frac{1}{1-\displaystyle\frac{f}{d}}=+\infty \).

Vậy \( \lim \limits_{d \to f^+} g(d)=+\infty \).

\textbf{Ý nghĩa:} Khi vật nằm tại tiêu điểm (\( a=OF=f \)) cho ảnh ở vô cùng.

b) \( \lim \limits_{d \to +\infty} g(d) =\lim \lim \limits_{d \to +\infty} \displaystyle\frac{df}{d-f}=\lim \limits_{d \to f^+} \displaystyle\frac{f}{1-\displaystyle\frac{f}{d}}=f\).

Vậy \( \lim \limits_{d \to +\infty} g(d) =f\).

\textbf{Ý nghĩa:} Khi vật ở rất xa (vô cực) cho ảnh tại tiêu điểm ảnh \( F' \) (\( d'=f \)).

}

a) Sau thời gian \( t \) phút, số lít nước trong hồ là \(6000+15t \) (lít).

Số gam muối trong số lít nước bơm vào là \( 30\cdot 15t=450t \) (gam).

Nồng độ muối của nước trong hồ sau \( t \) phút khi bắt đầu bơm là

\(C(t)=\displaystyle\frac{450t}{6000+15t}=\displaystyle\frac{30t}{400+t} \, \text{(gam/lít)}.\)

b) Khi \( t\to +\infty \), ta xét giới hạn

\(\lim \limits_{t \to +\infty} C(t)=\lim \limits_{t \to +\infty} \displaystyle\frac{30t}{400+t}=\lim \limits_{t \to +\infty} \displaystyle\frac{30}{\displaystyle\frac{400}{t}+1}=30.\)

Vậy khi bơm nước biểm vào hồ chứa, không giới hạn thời gian thì nồng độ muối trong hồ chứa chính là nồng độ muối của nước biển.

}