User:JaviPrieto/Limits

!x!!${\frac {\sin x}{x}}$

$\lim _{x\to p}f(x)=L,\,$

$\lim _{x\to p}f(x)=L,\$

$\lim _{x\to p^{+}}f(x)=L$

$\lim _{x\to p^{-}}f(x)=L$

$\lim _{x\to p}f(x)=L$

$\lim _{x\to p}f(x)=L$

$\lim _{x\to p}f(x)=L$

$\lim _{x\to \infty }f(x)=L,$

if and only if for all $\varepsilon >0$ there exists S > 0 such that $|f(x)-L|<\varepsilon$ whenever x > S.

$\lim _{x\to -\infty }f(x)=L,$

if and only if for all $\varepsilon >0$ there exists S < 0 such that $|f(x)-L|<\varepsilon$ whenever x < S.

$\lim _{x\to -\infty }e^{x}=0.\,$

$\lim _{x\to a}f(x)=\infty ,\,$

if and only if for all $\varepsilon >0$ there exists $\delta >0$ such that $f(x)>\varepsilon$ whenever $|x-a|<\delta$.

$\lim _{x\to \infty }f(x)=\infty ,\lim _{x\to a^{+}}f(x)=-\infty .\,$

$\lim _{x\to 0^{+}}\ln x=-\infty .\,$

$\lim _{x\to 0^{+}}{1 \over x}=\infty ,\lim _{x\to \infty }{1 \over x}=0.$

The complex plane with metric $d(x,y):=|x-y|$ is also a metric space. There are two different types of limits when the complex-valued functions are considered.

$\lim _{x\to p}f(x)=L$

if and only if for all e > 0 there exists a d > 0 such that for all real numbers x with $0<|x-p|<\delta$, then $|f(x)-L|<\varepsilon$.

$\lim _{(x,y)\to (p,q)}f(x,y)=L$

${\begin{matrix}\lim \limits _{x\to p}&(f(x)+g(x))&=&\lim \limits _{x\to p}f(x)+\lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x)-g(x))&=&\lim \limits _{x\to p}f(x)-\lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x)\cdot g(x))&=&\lim \limits _{x\to p}f(x)\cdot \lim \limits _{x\to p}g(x)\\\lim \limits _{x\to p}&(f(x)/g(x))&=&{\lim \limits _{x\to p}f(x)/\lim \limits _{x\to p}g(x)}\end{matrix}}$

$\lim _{y\to d}f(y)=e$, and $\lim _{x\to c}g(x)=d\Rightarrow \lim _{x\to c}f(g(x))=e$,

is not true. However, this "chain rule" does hold if, in addition, either f(d) = e (i. e. f is continuous at d) or g does not take the value d near c (i. e. there exists a $\delta >0$ such that if $0<|x-c|<\delta$ then $|g(x)-d|>0$).

• $\lim _{x\to 0}{\frac {\sin x}{x}}=1$
• $\lim _{x\to 0}{\frac {1-\cos x}{x}}=0$

$\sin x<x<\tan x.$

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

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

$\lim _{x\to 0}{\frac {1}{\cos x}}={\frac {1}{1}}=1$

$\lim _{x\to 0}{\frac {x}{\sin x}}=1$

$\lim _{x\to 0}{\frac {\sin x}{x}}=1$

• $\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$

$\lim _{x\to 0}{\frac {\sin(2x)}{\sin(3x)}}=\lim _{x\to 0}{\frac {2\cos(2x)}{3\cos(3x)}}={\frac {2\cdot 1}{3\cdot 1}}={\frac {2}{3}}.$ A short way to write the limit $\lim _{n\to \infty }\sum _{i=s}^{n}f(i)$ is $\sum _{i=s}^{\infty }f(i)$. A short way to write the limit $\lim _{x\to \infty }\int _{a}^{x}f(x)\;dx$ is $\int _{a}^{\infty }f(x)\;dx$. A short way to write the limit $\lim _{x\to -\infty }\int _{x}^{b}f(x)\;dx$ is $\int _{-\infty }^{b}f(x)\;dx$.