Limits 5

$\text{The technique is to use L'Hopital's rule to solve the limit.}$

$\text{Given,}$

$\displaystyle\lim_{x \to 0}\displaystyle\frac {x - \sin x}{x^{2}}$ = $\displaystyle\frac {0}{0}$

$\text{which is undefined when 0 is substitute for}$ $x .$

$\text{Thus, by applying L'Hopital's rule, the derivative of the numerator and the}$

$\text{denominator is taken separately, then, take the limit of the function.}$

$\displaystyle\lim_{x \to 0}\displaystyle\frac {x + (-\sin x)}{x^{2}}$ = $\displaystyle\lim_{x \to 0}\displaystyle\left [\frac {1 + (-\cos x)}{2x}\right ]$ = $\displaystyle\frac{1 + (-1)}{0}$ = $\displaystyle\frac {0}{0}$

$\text{which is still undefined when the limit is evaluated.}$

$\text{Apply the rule again to yield, and take the limit.}$

$\displaystyle\lim_{x \to 0}\displaystyle\frac{1 + (-\cos x)}{2x}$ = $\displaystyle\lim_{x \to 0} \displaystyle\left [\frac {\sin x}{2}\right ]$ = $\displaystyle\frac{0}{2}$ = $\displaystyle{0}$

$\text{Since, the}$ $\displaystyle\sin 0^\circ = 0$ $\text{and}$ $\displaystyle\frac {0}{n} = 0$ $\text{, where n is any integer}$ $\text{and n}$ $\neq 0$ . $\text{Therefore, the limit of the function is defined when x is approaches 0.}$