Các định lý và quy tắc sau đúng cho mọi trường hợp: $x\to {{x}_{0}},x\to x_{0}^{+},x\to x_{0}^{-},x\to +\infty ,x\to -\infty $.
Định lý 1: Giả sử $\underset{x\to {{x}_{0}}}{\mathop{\lim }}\,f\left( x \right)=L$và $\lim g\left( x \right)=M\left( L,M\in R \right)$. Khi đó
a)\[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \left[ {f\left( x \right) + g\left( x \right)} \right] = L + M;\]
b)\[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \left[ {f\left( x \right) - g\left( x \right)} \right] = L - M\];
c)\[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \left[ {f\left( x \right).g\left( x \right)} \right] = LM\];
Đặc biệt, nếu c là một hằng số thì \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \left[ {cf\left( x \right)} \right] = cL\];
d) Nếu $m\ne 0$ thì \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{L}{M}\].
Định lý 2: Giả sử \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} f\left( x \right) = L\], khi đó
a) \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \left| {f\left( x \right)} \right| = \left| L \right|\];
b) \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \sqrt[3]{{f\left( x \right)}} = \sqrt[3]{L};\]
c) Nếu $f\left( x \right)\ge 0$ với mọi $x\in J\backslash \left\{ {{x}_{0}} \right\}$ , trong đó J là một khoảng nào đó chứa ${{x}_{0}}$ thì $L\ge 0$ và \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \sqrt {f\left( x \right)} = \sqrt L \].
Định lý 3: Nếu \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \left| {f\left( x \right)} \right| = + \infty \] thì \[\mathop {\lim }\limits_{x \to {x_0}} {\mkern 1mu} \dfrac{1}{{f\left( x \right)}} = 0\].
Định lý 4: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
$ B=\underset{x\to \dfrac{\pi } 6 }{\mathop{\lim }}\,\dfrac{{{\sin }^ 2 }2 x -3\cos x }{\tan x}=\dfrac{{{\sin }^ 2 }\dfrac{\pi } 3 -3\cos \dfrac{\pi } 6 }{\tan \dfrac{\pi } 6 }$
$=\dfrac{\dfrac{3}{4} -\dfrac{3\sqrt{3} } 2 }{\dfrac{1}{{}\sqrt{3} }}=\dfrac{3}{4} \left( \sqrt{3} -6 \right) $
$ D=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt{3x+1}-2}{\sqrt[3]{3x+1}-2}=\dfrac{\sqrt{3+1}-2}{\sqrt[3]{3+1}-2}=\dfrac{0}{{}\sqrt[3] 4 -2}=0 $
Ta có: $ A=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{{ x ^ 2 }-x+1}{x+1}=\dfrac{1-1+1}{1+1}=\dfrac{1}{2} $ .
$ A=\underset{x\to -2}{\mathop{\lim }}\,\dfrac{x+1}{{ x ^ 2 }+x+4}=\dfrac{-2+1}{{{\left( -2 \right)}^ 2 }-2+4}=-\dfrac{1}{6} $
Ta có: $ C=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sqrt[3]{x+2}-x+1}{3x+1}=\sqrt[3] 2 +1 $
Do $ x\to -{ 1 ^ - }\Rightarrow \left| x+1 \right|=-(x+1) $ . Đáp số: $ \underset{x\to -{ 1 ^ - }}{\mathop{\lim }}\,\dfrac{{ x ^ 2 }+3x+2}{\left| x+1 \right|}=-1 $
$ C=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt{2{ x ^ 2 }-x+1}-\sqrt[3]{2x+3}}{3{ x ^ 2 }-2}=\sqrt{2} -\sqrt[3] 5 $
$ \underset{x\to { 2 ^ - }}{\mathop{\lim }}\,\dfrac{{ x ^ 2 }-4}{\sqrt{\left( { x ^ 4 }+1 \right)\left( 2-x \right)}}$
$=\underset{x\to { 2 ^ - }}{\mathop{\lim }}\,\dfrac{\left( x-2 \right)\left( x+2 \right)}{\sqrt{\left( { x ^ 4 }+1 \right)\left( 2-x \right)}}$
$=\underset{x\to { 2 ^ - }}{\mathop{\lim }}\,\dfrac{-\sqrt{2-x}\left( x+2 \right)}{\sqrt{{ x ^ 4 }+1}}=0 $
Ta có: $ D=\underset{x\to 1}{\mathop{\lim }}\,\dfrac{\sqrt[3]{7x+1}+1}{x-2}=\dfrac{\sqrt[3] 8 +1}{1-2}=-3 $ .
Ta có $ B=\underset{x\to \dfrac{\pi } 6 }{\mathop{\lim }}\,\dfrac{2\tan x+1}{\sin x+1}=\dfrac{2\tan \dfrac{\pi } 6 +1}{\sin \dfrac{\pi } 6 +1}=\dfrac{4\sqrt{3} +6} 9 $ .
Ta có $ \underset{x\to { 2 ^ + }}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to { 2 ^ + }}{\mathop{\lim }}\,\left( { x ^ 2 }-3 \right)=1 $
$ \underset{x\to { 2 ^ - }}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to { 2 ^ - }}{\mathop{\lim }}\,\left( x-1 \right)=1 $
Vì $ \underset{x\to { 2 ^ + }}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to { 2 ^ - }}{\mathop{\lim }}\,f\left( x \right)=1 $ nên $ \underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)=1 $ .
$ \underset{x\to -\infty }{\mathop{\lim }}\,\left( { x ^ 2 }+x-1 \right)=\underset{x\to -\infty }{\mathop{\lim }}\,{ x ^ 2 }\left( 1+\dfrac{1}{x} -\dfrac{1}{{}{ x ^ 2 }} \right)=+\infty $