1. ## calculus limits

Hi mathsforum

I needed some help with the following:

Find

1. $\displaystyle \lim_{x \to 0} \left\{ \frac{1-2sin^2x-cos^3x}{5x^2}\right\}$

2. $\displaystyle \lim_{x \to 0} \left\{ \frac{x-sinx}{x-tanx}\right\}$

thank you.

1. $\displaystyle \lim_{x \to 0} \left\{ \frac{1-2sin^2x-cos^3x}{5x^2}\right\}$
This is of the form $\displaystyle \frac{0}{0}$ so we may use L'Hopital's rule:
$\displaystyle \lim_{x \to 0}\frac{1-2sin^2(x)-cos^3(x)}{5x^2}$

$\displaystyle = \lim_{x \to 0}\frac{-4sin(x)cos(x) + 3sin(x)cos^2(x)}{10x}$

This is still of the form $\displaystyle \frac{0}{0}$ so we use L'Hopital's rule again:
$\displaystyle = \lim_{x \to 0}\frac{-4cos^2(x) + 4sin^2(x) + 3cos^3(x) - 3sin^2(x)cos(x)}{10} = \frac{-4 + 3}{10} = -\frac{1}{10}$

-Dan

2. $\displaystyle \lim_{x \to 0} \left\{ \frac{x-sinx}{x-tanx}\right\}$
Again using L'Hopital's rule:
$\displaystyle \lim_{x \to 0}\frac{x-sinx}{x-tanx}$

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

Again apply L'Hopital's rule:
$\displaystyle = \lim_{x \to 0}\frac{sin(x)}{-2tan^3(x) - 2tan(x)}$

Again:
$\displaystyle = \lim_{x \to 0}\frac{cos(x)}{-6tan^4(x) - 8tan^2(x) - 2} = \frac{1}{-2} = - \frac{1}{2}$

-Dan