Evaluate $\lim_{x\to 0^+}x^{\sin x}$ without L’Hospital $$\lim_{x\to 0^+}x^{\sin x}$$
I know it can be easily solved using L’Hospital’s Theorem, but I would like to see another way.
Thanks in advance
 A: $$x^{\sin x}=e^{\sin x\ln x}=e^{\frac{\sin x}{x}x\ln x}\to e^0=1$$
since: $$x \log x \to 0$$ $$\frac{\sin x}{x}\to1$$
A: We assume that the domain here is $\mathbb{R}_{> 0}$.
Write $x^{\sin x} = \exp\left(\log x \sin x\right)$
By the continuity of $\exp$, it follows that 
$$\lim_{x \to 0} \exp\left(\log x \sin x\right) = \exp\left(\lim_{x \to 0} (\log x \sin x)\right)$$
and we may note that $$\lim_{x \to 0} \log x \sin x = \lim_{x \to 0} x\log x \frac{\sin x}{x} =\lim_{x \to 0} x\log x = 0$$
which implies that the original limit is $1$. 
The limit $\lim_{x \to 0} x \log x = 0$ can be shown by letting $x = e^{-u}$ and noting that $$\lim_{x \to 0} x\log x = \lim_{u \to \infty} -ue^{-u}$$
and $$0 \leq \lim_{u \to \infty} -ue^{-u} = \lim_{u \to \infty} \frac{-u}{e^u} \leq \lim_{u \to \infty}\frac{-u}{1 + u + \frac{u^2}{2}} = 0 $$
which implies $\lim_{u \to \infty} -ue^{-u} = 0$. Note that we used the second order Taylor expansion of $\exp$ in the above inequality. 

Warning: 
A natural approach to solving this problem, as has been done by others, is writing $x^{\sin x}= x^\frac{x\sin x}{x}$
or, equivalently, $x^{\sin x}= (x^x)^\frac{\sin x}{x}$
and using the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to conclude that $\lim_{x \to 0} x^{\sin x} = \lim_{x \to 0} x^x = 1$.
Although perhaps this argument yields the correct answer for this "well-behaved" example, in general this sort of technique is spurious.  
A counterexample I like is $$\lim_{x \to 0} \frac{\exp \left(\frac{1}{\sin^2(x)}\right)}{\exp\left(\frac{1}{x^2}\right)}$$
Naively, we might note that 
$$\lim_{x \to 0} \frac{\exp \left(\frac{1}{\sin^2 (x)}\right)}{\exp\left(\frac{1}{x^2}\right)} = \lim_{x \to 0} \frac{\exp \left(\frac{1}{x^2} \cdot \frac{x^2}{\sin^2(x)}\right)}{\exp\left(\frac{1}{x^2}\right)} $$
and since in the numerator $\frac{x^2}{\sin^2(x)} \to 1$ we'd be inclined to think that the limit is  $$\lim_{x \to 0} \frac{\exp \left(\frac{1}{x^2} \cdot \frac{x^2}{\sin^2(x)}\right)}{\exp\left(\frac{1}{x^2}\right)} = \lim_{x \to 0} \frac{\exp \left(\frac{1}{x^2}\right)}{\exp\left(\frac{1}{x^2}\right)} = 1$$
This, however, is incorrect. The limit is, in fact, 
$$\lim_{x \to 0} \frac{\exp \left(\frac{1}{\sin^2(x)}\right)}{\exp\left(\frac{1}{x^2}\right)} = \lim_{x \to 0}  \exp\left(\frac{1}{\sin^2(x)} - \frac{1}{x^2}\right) = \lim_{x \to 0}  \exp\left(\frac{x^2 - \sin^2(x)}{x^2\sin^2(x)}\right)$$
hence this amounts to computing $$\lim_{x \to 0} \frac{x^2 - \sin^2(x)}{x^2\sin^2(x)} \stackrel{\sin x \sim x}{=} \lim_{x \to 0} \frac{x^2  + \frac{1}{2}(\cos(2x) - 1)}{x^4} = \lim_{x \to 0} \frac{2x^2  + \cos(2x) - 1}{2x^4} =  \lim_{x \to 0} \frac{2x^2  + (-2x^2 + \frac{2x^4}{3} + O(x^6))}{2x^4} =  \frac{1}{3}$$ 
so in fact $$\lim_{x \to 0} \frac{\exp \left(\frac{1}{\sin^2(x)}\right)}{\exp\left(\frac{1}{x^2}\right)} = e^{\frac 13} \neq 1$$
A: Hint :
Let's write our function using the known exponent power rule $a^x=e^{x\ln(a)}$;
$$x^{\sin(x)} = e^{\sin(x)\ln(x)}$$
A: Multiply and divide the exponent by $x$ to get
$$\large x^{\frac{x \sin(x)}{x}}$$
Due to the well known limit $sin x / x \to 1$ hence
$$x^x$$
Which then is
$$\boxed{1}$$
A: HINT:
Note that $$x^{\sin(x)}=(x^x)^{\frac{\sin(x)}{x}}$$
A: If you want more than the limit, consider
$$A=x^{\sin(x)}\implies \log(A)=\sin(x)\, \log(x)$$ Now, use the Taylor expansion of $\sin(x)$ to get
$$\log(A)=x \log (x)-\frac{1}{6} x^3 \log (x)+O\left(x^4\right)$$ Now, continuing with Taylor
$$A=e^{\log(A)}=1+x \log (x)+\frac{1}{2} x^2 \log ^2(x)+\frac{1}{6} x^3 \log (x) \left(\log
   ^2(x)-1\right)+O\left(x^4\right)$$
Let us $x=\frac \pi 6$; the above approximation gives $0.727603$ while $\sqrt {\frac \pi 6}\approx 0.723601$.
