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disqus: abhyas29
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# Limit

###### tags: `Mathematics`

[TOC]

[All Mathematics Formula by Abhyas here](/@abhyas/maths_formula)

Please see [README](/@abhyas/maths_formula#README) if this is the first time you are here.

## Method of Limits

### Direct Substitution

Put value of x if no indefinite from occurs.
$\lim_{x \to 2} \frac{x+2}{x^2+4}=\frac{2+2}{2^2+4}=\frac{1}{2}$

Limit is only solved for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form. Else use direct substation.

### Factorization Method

If elimination of indefinite form possible using factorization then cancel it out.
$\lim_{x \to 3} \frac{x^2-5x+6}{x-3}$
at $x=3$, $\frac{3^2-5 \times 3+6}{3-3}=\frac{0}{0}$ is undefined. So factorize as:
$1$

*Better use L'Hôpital's rule*

### Rationalization Method

Multiply numerator and denominator by rationalizing factor.
$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x}$
at $x=0$, $\frac{0}{0}$ is undefined. So rationalize as:
$\lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x} \times \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$
$=\lim_{x \to 0} \frac{2}{(\sqrt{1+x}+\sqrt{1-x})}$
at $x=0$, $\frac{2}{\sqrt{1+0}+\sqrt{1-0}}=1$

*Better use L'Hôpital's rule*

### limit when $x \to \infty$

### L'Hôpital's rule

**Required Condition:** $\lim_{x \to a}\frac{f(x)}{g(x)}=\frac{f(a)}{g(a)}=\frac{0}{0}$ or $\frac{\infty}{\infty}$
and according to the rule, differentiate numerator and denominator *separately*
$\lim_{x \to a} \frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}$
## Licensing and Links

[All Mathematics Formula by Abhays here](https://hackmd.io/@abhyas/maths_formula)

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