a. Áp dụng công thức L'Hospital:

\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{1-x}}{\sqrt[3]{x+1}-\sqrt{1-x}}=\lim\limits_{x\to 0}\frac{\frac{1}{2}(x+1)^{\frac{-1}{2}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}{\frac{1}{3}(x+1)^{\frac{-2}{3}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}=\frac{1}{\frac{5}{6}}=\frac{6}{5}\)

b.

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

c. Áp dụng quy tắc L'Hospital:

\(\lim\limits_{x\to +\infty}\frac{x^4-x^3+11}{2x-7}=\lim\limits_{x\to +\infty}\frac{4x^3-3x^2}{2}=+\infty \)

d.

\(\lim\limits_{x\to 5}\frac{7}{(x-1)^2}.\frac{2x+1}{2x-3}=\frac{7}{(5-1)^2}.\frac{2.5+11}{2.5-3}=\frac{11}{16}\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)

a. \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)

b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)

Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)

\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)

\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)

Ta có : (...) = \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)-\left[\sqrt[3]{x^2+7}-\left(x+1\right)\right]}{x^2-1}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{5-x^3-\left(x+1\right)^2}{\left(\sqrt{5-x^3}+x+1\right)\left(x^2-1\right)}\)  

\(=\lim\limits_{x\rightarrow1}\dfrac{-x^3-x^2-2x+4}{...}\)  \(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)\left(x-1\right)}{...}\)   

= \(\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)}{\left(x+1\right)\left(\sqrt{5-x^3}+x+1\right)}=\dfrac{-7}{8}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{x^2+7-x^3-3x^2-3x-1}{\left(x^2-1\right)\left[\sqrt[3]{\left(x+7\right)^2}+\left(x+1\right)\sqrt[3]{x^2+7}+\left(x+1\right)^2\right]}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)\left(x-1\right)}{...}\)  

\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)}{\left(x+1\right)\left[\sqrt[3]{\left(x^2+7\right)^2}+\sqrt[3]{x^2+7}\left(x+1\right)+\left(x+1\right)^2\right]}\)

\(=\dfrac{-\left(1+3+6\right)}{\left(1+1\right)\left(4+2.2+4\right)}=\dfrac{-5}{12}\)

Suy ra : \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=\dfrac{-7}{8}+\dfrac{5}{12}=\dfrac{-11}{24}\)

Cách 2 : Tách : \(\sqrt{5-x^3}-2-\left(\sqrt[3]{x^2+7}-2\right)\) -> Dùng liên hợp 

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)

\(=\dfrac{23}{12}\)

a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)

Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:

\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)

\(=-1\)

L'Hospital đi em

lim đề bài cho = \(\lim\limits_{x->\dfrac{\pi}{4}}\dfrac{-\sqrt{2}sinx}{\sqrt{2}cosx}\)

Thay x vào là xong

làm bài lim xem nào :))) 

P/s Sở Kiều :))

Hướng làm thôi chứ gõ công thức lâu vl

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+x+1}-\sqrt[3]{x^3+1}}{x}\) 

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+x+1}-1}{x}+\dfrac{1-\sqrt[3]{x^3+1}}{x}\)

Đến đây liên hợp là xong phần của bạn đó ;) 

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)

Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)