Limits of functions | 9math

Limits of functions

Posted February 2nd, 2008 by Isoscel

We list some useful limits .

For $x\in (-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2})$ we have

$|\sin x| <|x|<| \tan x|=\frac{|\sin x|}{\cos x}$

$\frac{\cos x}{|\sin x|}<\frac{1}{| x|}<\frac{1}{|\sin x|}$

$\cos x<|\frac{\sin x}{ x}|<1$

$\cos x<\frac{\sin x}{ x}<1$

$-1<-\frac{\sin x}{x}<-\cos x$

$0<1-\frac{\sin x}{x}<1-\cos x$

$|\frac{\sin x}{x}-1|<1-\cos x$

as

$\lim_{x\to 0}1-\cos x=0$

$\lim_{x\to 0}\frac{\sin x}{x}=1\:;\lim_{x\to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}$

$\lim_{x\to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2}$

$\lim_{x\to 0}\frac{e^x-1}{x}=1\\:\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$

$\lim_{x\to 0}\frac{\ln (1+x)}{x}=1\:;\lim_{x\to 1}\frac{\ln x}{x-1}=1$

$\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a$

$a>0\:\:\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$

$\mathop{\lim}\limits_{x \to \infty}(a^\frac{1}{x}-1)x =\lim_{y\to 0}\frac{a^y-1}{y}=\lim_{y\to 0}\frac{e^{\ln a^y}-1}{y}=\lim_{y\to 0}\frac{e^{y\ln a}-1}{y}=\ln a$

$\lim_{x \to 0}(1+x)^{\frac{1}{x}}=e$

$\lim_{x \to 0}(1-x)^{\frac{1}{x}}=\frac{1}{e}$

$\lim_{x \to 0}(1+\sin x)^{\frac{1}{x}}=e$

$\lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}=\lim_{x \to 0}(1-(1-\cos x))^{\frac{1}{1-\cos x}\frac{1-\cos x}{x^2}}=e^{-\frac{1}{2}}$

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Comments

This is a very useful article.

On December 2nd, 2009 alpha says:

So many many thanks.

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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