# Limits at Infinity

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Definition of Limits at infinity
Let 'L' be any real number.
1)$\lim_{x \rightarrow \infty}f(x) = L$ means that for each $\epsilon$ > 0 there exists M > 0 such that
|f(x) - L| < $\epsilon$ whenever x > M.

2) $\lim_{x \rightarrow - \infty}f(x) = L$ means that for each $\epsilon$ > 0 there exists N < 0 such that |f(x) - L| < $\epsilon$ whenever x < N.
Horizontal asymptotes
The line is a horizontal asymptote of the graph of when
$\lim_{x \rightarrow -\infty}f(x) = L$ OR $\lim_{x \rightarrow \infty}f(x)$ = L

Vertical asymptotes at x = a
The line is a vertical asymptote of the graph of when
$\lim_{x \rightarrow a}f(x) = \infty$ OR $\lim_{x \rightarrow \infty}f(x) = -\infty$

STEPS FOR FINDING LIMITS AT $\pm\infty$ OF RATIONAL FUNCTIONS
1) If the degree of the numerator < the degree of the denominator, then the limit of the rational function is 0.
2) If the degree of the numerator = the degree of the denominator, then the limit of the rational function is the ratio of the leading coefficients.
3) If the degree of the numerator > the degree of the denominator, then the limit of the rational function does not exist.

## Examples on Limits at Infinity

Example 1 : Find the limit of $\lim_{x \rightarrow \infty}\frac{2x-1}{3x +5}$

Solution : $\lim_{x \rightarrow \infty}\frac{2x-1}{3x +5}$

Take x as a common factor from numerator and denominator
= $\lim_{x \rightarrow \infty}\frac{x(2-\frac{1}{x})}{x(3 +\frac{5}{x})}$

= $\lim_{x \rightarrow \infty}\frac{2-\frac{1}{x}}{3 +\frac{5}{x}}$

( $\lim_{x \rightarrow \infty}\frac{1}{x} = 0)$

$\lim_{x \rightarrow \infty}\frac{2x-1}{3x +5} = \frac{2}{3}$

Example 2 : Find the limit of $\lim_{x \rightarrow \infty}\frac{2x^{2}+4 }{x^{2} -5x -1}$

Solution : $\lim_{x \rightarrow \infty}\frac{2x^{2}+4 }{x^{2} -5x -1}$

Take $x^{2}$ as a common factor from numerator and denominator
= $\lim_{x \rightarrow \infty}\frac{x^{2}(2+\frac{4}{x^{2}})}{x^{2}(1 -\frac{5}{x}-\frac{1}{x^{2}})}$

$\lim_{x \rightarrow \infty}\frac{(2+\frac{4}{x^{2}})}{(1 -\frac{5}{x}-\frac{1}{x^{2}})}$ ( $\lim_{x \rightarrow \infty}\frac{1}{x} = 0$ and $\lim_{x \rightarrow \infty}\frac{1}{x^{2}} = 0$)

$\lim_{x \rightarrow \infty}\frac{2x^{2}+4 }{x^{2} -5x -1}$ =2