Limit with L'Hospital with infinite indeterminate formats I'm trying to find the limit:
$$\large \lim_{x\to0}(\sin x)^x$$
Whst I did was apply L'Hospital Rule:
$$\large \text{let }y =(\sin x)^x\implies
\ln y=x\ln\sin x$$
$$\large 
\lim_{x\to0}\ln y
=
\lim_{x\to0} x\ln\sin x
=
\lim_{x\to0}\frac x{\frac1{\ln\sin x}}
=
\lim_{x\to0} \frac1{\frac {-\cos x}{(\ln\sin x)^2\sin x}}
=
\lim_{x\to0} (-\tan x )(\ln\sin x)^2
=
\lim_{x\to0} \frac{(\ln\sin x)^2}{\frac1{(-\tan x )}}
=
...
$$
Ultimately it keeps on going, please help me.
 A: if you have $\frac{\log \sin x}{\frac{1}{x}}$ and apply L'Hospital's rule, the log will disappear, so you got to fix it
A: HINT:
$$\lim_{x\to0^+}\frac{\ln\sin x}{\dfrac1x}=\lim_{x\to0^+}\frac{\dfrac{\cos x}{\sin x}}{-\dfrac1{x^2}}=-\lim_{x\to0}x\cdot \lim_{x\to0}\cos x\cdot\frac1{\lim_{x\to0}\dfrac{\sin x}x}$$
Hope you can take it home from here?
A: Hint: $$\sin x\sim_0 \, x \ \ \Rightarrow \ \ \lim_{x\to0}\:(\sin x)^x=\lim_{x\to0}\:x^x.$$
A: Whgen evaluating the limit of a product with l'Hopital's rule, I'd advise to put in the denominator the simplest factor. So:
$$\begin{align}
\lim_{x\to0} x\ln\sin x&=\lim_{x\to0}\frac{\ln\sin x}{1/x}\\
&=-\lim_{x\to0}x^2\cot x\\
&=\lim_{x\to0}\frac{x^2}{\tan x}\\
&=\lim_{x\to0}\frac{2x}{1+\tan^2x}\\
&=0
\end{align}$$
A: Let the desired limit be $L$ then we have $$\begin{aligned}\log L &= \log\left(\lim_{x \to 0^{+}}(\sin x)^{x}\right)\\
&= \lim_{x \to 0^{+}}\log(\sin x)^{x}\text{ (by continuity of log)}\\
&= \lim_{x \to 0^{+}}x\log(\sin x)\\
&= \lim_{x \to 0^{+}}x\log\left(\frac{\sin x}{x}\cdot x\right)\\
&= \lim_{x \to 0^{+}}x\log\left(\frac{\sin x}{x}\right) + x\log x\\
&= 0\cdot \log 1 + \lim_{x \to 0^{+}}x\log x\\
&= -\lim_{y \to \infty}\frac{\log y}{y}\text{ (putting }y = 1/x)\\
&= 0\end{aligned}$$ so that the desired limit $L = e^{0} = 1$. Here we have used two standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\,\lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$$ for any $a > 0$. We really don't need powerful tools like L'Hospital for such common problems.
