Tìm giới hạn : \(L=\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{\cos x}-\sqrt[5]{\cos x}}{\sin^2x}\)
Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :
\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)
\(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+-.+y+1\right)}=-\frac{1}{40}\)
Tính giới hạn hàm số :
\(\lim\limits_{x\rightarrow10}\frac{lgx-1}{x-10}\)
\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\tan x}\)
\(\lim\limits_{x\rightarrow0}\frac{e^x-1}{\sqrt{x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\frac{e^{5x+3}-e^3}{2x}\)
\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{2x}\)
\(\lim\limits_{x\rightarrow0}\frac{e^x-e^{-x}}{\sin x}\)
\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)
\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}\)
\(\lim\limits_{x\rightarrow e}\frac{\ln x-1}{x-e}\)
tính Lim(x-->0)\(\frac{1}{\sqrt[3]{\left(x+1\right)^2+\sqrt[3]{x+1}+1}}\)
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