\(L=\lim\limits_{x\rightarrow0}\frac{e^x-1}{\sqrt{x+1}-1}=\lim\limits_{x\rightarrow0}\frac{\left(e^x-1\right)\left(\sqrt{x+1}-1\right)}{x}=\lim\limits_{x\rightarrow0}\left[\frac{e^x-1}{x}.\left(\sqrt{x+1}-1\right)\right]=1.0=0\)

\(L=\lim\limits_{x\rightarrow0}\frac{e^{5x+3}-e^3}{2x}=\lim\limits_{x\rightarrow0}\left(\frac{e^{5x}-1}{5x.\frac{2}{5}}.e^3\right)=\lim\limits_{x\rightarrow0}\left(\frac{e^{5x}-1}{5x}.\frac{5e^3}{2}\right)=1.\frac{5e^3}{2}=\frac{5e^3}{2}\)

Bài 1:

\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)

Bài 2:

\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)

\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)

\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)

Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)

Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)

\(L=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{2x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{x^3.\frac{2}{x^2}}=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+x^3\right)}{x^3}.\frac{x^3}{2}\right]=1.0=0\)

\(L=\lim\limits_{x\rightarrow0}\frac{\ln x-1}{\tan x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\frac{\sin x}{\cos x}}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{2x.\frac{\sin x}{x}.\frac{1}{2\cos x}}\)

   \(=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+2x\right)}{2x}.\frac{1}{\frac{\sin x}{x}}.2\cos x\right]=1.\frac{1}{1}.2.1=2\)

Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :

\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)

    \(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+.....+y+1\right)}=-\frac{1}{40}\)

Đặt \(t=x-e\Rightarrow\begin{cases}x=t+e\\x\rightarrow e;t\rightarrow0\end{cases}\)

\(\Rightarrow L=\lim\limits_{t\rightarrow0}\frac{\ln\left(t+e\right)-\ln e}{t}=\lim\limits_{t\rightarrow0}\frac{\ln\left(\frac{t+e}{e}\right)}{t}=\lim\limits_{t\rightarrow0}\left[\frac{\ln\left(1+\frac{t}{e}\right)}{\frac{t}{e}}\right]=\frac{1}{e}\)

Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:

\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)

\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)

\(c=0.sin\frac{1}{2}=0\)

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