1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\) 

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)

2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0

3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)

\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)

4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)

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

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

Vì \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ - }} {\rm{ }}f\left( x \right)\) nên không tồn tại \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\).

D

D

D

a) Áp dụng giới hạn một bên thường dùng, ta có : \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}} =  + \infty \)

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

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

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} =  - \infty \)

a) Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì, \({x_n} >  - 1\) và \({x_n} \to  - 1\). Khi đó \(f\left( {{x_n}} \right) = x_n^2 + 2\)

Ta có: \(\lim f\left( {{x_n}} \right) = \lim \left( {x_n^2 + 2} \right) = \lim x_n^2 + \lim 2 = {\left( { - 1} \right)^2} + 2 = 3\)

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

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì, \({x_n} <  - 1\) và \({x_n} \to  - 1\). Khi đó \(f\left( {{x_n}} \right) = 1 - 2{x_n}\).

Ta có: \(\lim f\left( {{x_n}} \right) = \lim \left( {1 - 2{x_n}} \right) = \lim 1 - \lim \left( {2{x_n}} \right) = \lim 1 - 2\lim {x_n} = 1 - 2.\left( { - 1} \right) = 3\)

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

b) Vì \(\mathop {\lim }\limits_{x \to  - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to  - {1^ - }} {\rm{ }}f\left( x \right) = 3\) nên \(\mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = 3\).

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

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

b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) = \mathop {\lim }\limits_{x \to 4} x + \mathop {\lim }\limits_{x \to 4} 4 = 4 + 4 = 8\)

c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)

                                         \( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)

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

\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+2}-2}{x-2}\)

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

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\cdot\sqrt{1+\dfrac{2}{x^2}}-2}{x\left(1-\dfrac{2}{x}\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{2}{x^2}}-\dfrac{2}{x}}{1-\dfrac{2}{x}}=\dfrac{\sqrt{1+0}-0}{1-0}=\dfrac{1}{1}=1\)

Vì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = 3 \ne \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 5\) nên không tồn tại giới hạn \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\)

Để giới hạn đã cho hữu hạn

\(\Rightarrow\sqrt{x^2+mx-m-3}-x=0\) có nghiệm \(x=4\)

\(\Rightarrow\sqrt{16+4m-m-3}-4=0\)

\(\Rightarrow\sqrt{3m+13}=4\Rightarrow m=1\)

Khi đó:

 \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{x^2+x-4}-x}{x^2-5x+4}=\lim\limits_{x\rightarrow4}\dfrac{x-4}{\left(x-1\right)\left(x-4\right)\left(\sqrt{x^2+x-4}+x\right)}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{1}{\left(x-1\right)\left(\sqrt{x^2+x-4}+x\right)}=\dfrac{1}{3\left(\sqrt{4^2+4-4}+4\right)}=\dfrac{1}{24}\)