Tính giới hạn hàm số :
\(\lim\limits_{x\rightarrow10}\frac{lgx-1}{x-10}\)
Đặt \(t=x-10\Rightarrow\begin{cases}x=t+10\\x\rightarrow t;t\rightarrow0\end{cases}\)
\(\Rightarrow L=\lim\limits_{t\rightarrow0}\frac{lg\left(t+10\right)-lg10}{t}=\lim\limits_{t\rightarrow0}\frac{lg\left(\frac{t+10}{10}\right)}{t}=\lim\limits_{t\rightarrow0}\left[\frac{lg\left(1+\frac{t}{10}\right)}{\frac{t}{10}}.\frac{1}{10}\right]=\frac{1}{10}\)
\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\tan x}\)
\(\lim\limits_{x\rightarrow0}\frac{e^x-1}{\sqrt{x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\frac{e^{5x+3}-e^3}{2x}\)
\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{2x}\)
\(\lim\limits_{x\rightarrow0}\frac{e^x-e^{-x}}{\sin x}\)
\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)
\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}\)
\(\lim\limits_{x\rightarrow e}\frac{\ln x-1}{x-e}\)
tính Lim(x-->0)\(\frac{1}{\sqrt[3]{\left(x+1\right)^2+\sqrt[3]{x+1}+1}}\)
Cho \(y=\sin\left(\ln x\right)+\cos\left(\ln x\right)\). Chứng minh hệ thức : \(y+xy'+x^2y"=0\)
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