# Infinite Limits

Definition of infinite limits : Let 'f' be a function that is defined at every real number in some open interval containing 'c' from both sides except at 'c' itself then
$\lim_{x->c}f(x) = \infty$

means that for each M > 0 there exists a $\delta$ > 0 such that f(x) > M whenever
0 < |x - c| < $\delta$ So,

$\lim_{x->c}f(x) = -\infty$
means that for each N < 0 there exists a $\delta$ > 0 such that f(x) < N whenever
0 < |x - c| < $\delta$ So,
To define infinite limit from the left replace 0 < |x - c| < $\delta$ by
c - $\delta$ < x < c
To define the infinite-limit from the right replace 0 < |x - c| < $\delta$
by c < x < c + $\delta$

## Examples on Infinite Limits

1) Evaluate the $\lim_{x->-1^{+}}\frac{1}{x+1}$

Solution : To find this limit test a few values that are close to -1 as you approach from the right.
x = -0.9
$\frac{1}{-0.9 + 1}$ = 10

x = - 0.99
$\frac{1}{-0.99 + 1}$ = 100

x = - 0.999
$\frac{1}{-0.999 + 1}$ = 1000

From the results above we see that the limit increases without bound as we approach -1 from the right.

$\lim_{x->-1^{+}}\frac{1}{x+1}$ = $\infty$

2) Evaluate the $\lim_{x->0^{+}}\frac{2}{sin(x)}$ where x is measured in radian.

Solution : To find this limit test a few values that are close to 0 as you approach from the right.
x = 0.1
$\frac{2}{sin(0.1)} \approx$ 20

x = 0.01
$\frac{2}{sin(0.01} \approx$ 200

x = 0.001
$\frac{2}{sin(0.001} \approx$ 2000

From the results above we see that the limit increases without bound as we approach -1 from the right.

$\lim_{x->0^{+}}\frac{2}{sin(x)}$ = $\infty$