Cho m là số nguyên dương. Tìm giới hạn sau :
\(L_m=\lim\limits_{x\rightarrow1}\left(\frac{m}{1-x^m}-\frac{1}{1-x}\right)\)
Ta có \(L_m=\lim\limits_{x\rightarrow1}\left(\frac{m-\left(1+x+x^2+-.+x^{m-1}\right)}{1-x^m}\right)\)
\(=\lim\limits_{x\rightarrow1}\frac{\left(1-x\right)+\left(1-x^2\right)+-.+\left(1-x^{m-1}\right)}{1-x^m}\)
\(=\lim\limits_{x\rightarrow1}\frac{\left(1-x\right)\left[1+\left(1+x\right)+-.+\left(1+x+x^2+-.+x^{m-2}\right)\right]}{\left(1-x\right)\left(1+x+x^2+-.+x^{m-1}\right)}\)
\(=\frac{1+2+3+-+\left(m-1\right)}{m}=\frac{\left(m-1\right)m}{2m}=\frac{m-1}{2}\)
Tìm giới hạn :
\(L=\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{\left(x+a_1\right)\left(x+a_1\right)\left(x+a_1\right)}-x\right)\)
Tìm giới hạn : \(L=\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{\cos x}-\sqrt[5]{\cos x}}{\sin^2x}\)
Tính giới hạn hàm số :
\(\lim\limits_{x\rightarrow10}\frac{lgx-1}{x-10}\)
\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\tan x}\)
\(\lim\limits_{x\rightarrow0}\frac{e^x-1}{\sqrt{x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\frac{e^{5x+3}-e^3}{2x}\)
\(\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{2x}\)
\(\lim\limits_{x\rightarrow0}\frac{e^x-e^{-x}}{\sin x}\)
\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)
\(\lim\limits_{x\rightarrow+\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}\)
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