# What is the limit of  (1-cos(x))/sin(x^2)  as x approaches 0?

Jul 8, 2016

${\lim}_{x \to 0} \frac{1 - \cos \left(x\right)}{\sin \left({x}^{2}\right)} = \frac{0}{0}$ (an indeterminate form), thus we can apply

L'Hospital's rule, since substituting $x = 0$ yields $\frac{0}{0}$.

In this case, applying L'Hospital's rule we get

${\lim}_{x \to 0} \frac{\sin \left(x\right)}{2 x \cdot \cos \left({x}^{2}\right)} \to \frac{0}{0}$, which also yields an indeterminate form,

thus we apply it again:

${\lim}_{x \to 0} \frac{\cos \left(x\right)}{2 \cos \left({x}^{2}\right) - 4 {x}^{2} \cdot \sin \left({x}^{2}\right)}$

=lim_(x->0) (cos(x))/(2(cos(x^2)-2x^2 * sin(x^2))

=cos(0)/(2(cos(0)-2(0)(sin(0))

$= \frac{1}{2 \left(1 - 0\right)} = \frac{1}{2}$

#### Explanation:

L'Hospital's rule states that if function(s) $f$ and $g$ are differentiable on an open interval $I$ except at $c$ contained in $I$, then if

${\lim}_{x \to c} f \left(x\right) = {\lim}_{x \to c} g \left(x\right) = 0$ or ± ∞,

g'(x) ≠ 0 for all $x \in I$ with x ≠ c, and

${\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists, then

${\lim}_{x \to c} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

See more examples here.