# What is the limit as x turns to 0 of x^sinx?

May 11, 2018

${\lim}_{x \to 0} {x}^{\sin x} = 1$

#### Explanation:

If we try to directly substitute $x = 0$ into ${\lim}_{x \to 0} {x}^{\sin x}$, we end up with ${0}^{0}$, an indeterminate form. Thus, this is a case in which L'Hospital's Rule is useful.

Consider that our function is $y = {x}^{\sin x}$. Taking the log of both sides gives $\ln y = \ln \left({x}^{\sin x}\right) = \sin x \ln x$. L'Hospital's Rule generally works only on quotients, so we note that $\sin x = \frac{1}{\csc x}$.

This gives ${\lim}_{x \to 0} \ln y = {\lim}_{x \to 0} \sin x \ln x = {\lim}_{x \to 0} \ln \frac{x}{\frac{1}{\sin} x}$. At this point we can actually apply L'Hospital's rule, giving: lim_(x->0) lnx/(1/sinx) = lim_(x->0) (1/x)/(-cosx/(sin^2x).

We continue to rearrange and apply L'Hospital's rule as necessary. I will omit the ${\lim}_{x \to 0}$ for now.

$\frac{\frac{1}{x}}{\frac{- \cos x}{{\sin}^{2} x}} = {\sin}^{2} \frac{x}{- x \cos x} \implies \left(L ' H\right) \implies \frac{2 \sin x \cos x}{- \cos x + x \sin x}$

At this point, we have a determinate form.

lim_(x->0) (2sinxcosx)/(xsinx - cosx) = (2sin(0)cos(0))/((0)(sin(0)) - cos(0)
$= \frac{0}{0 - 1} = \frac{0}{- 1} = 0$

So we've found that ${\lim}_{x \to 0} \ln y = 0$. Note that $y = {e}^{\ln y}$. Thus our original limit can be written ${\lim}_{x \to 0} {e}^{\ln y}$. Note that $e$ is just a constant, so we can simply plug in the limit we found for $\ln y$ as $x \to 0$, giving:

${\lim}_{x \to 0} {x}^{\sin x} = {\lim}_{x \to 0} {e}^{\ln y} = {e}^{0} = 1$.