Compute the limit $\lim_{x\to^+}x^{\tan(x^2)}$

Question

I've been struggling with evaluating this limit for quite a while now: $$\lim_{x\rightarrow0^+}x^{\tan(x^2)}$$ I have tried expressing the function above as $$e^{\tan(x^2)\ln x}$$ and working my way from here, but I can't seem to find a way to get rid of the indetermination. While Wolfram Alpha and Symbolab do have an answer, neither of them seem to be capable of providing the steps to solve it.

Answer 1

$$\lim_{x\to0} \tan(x^2)\ln(x)=\lim_{x\to0} \frac{\tan(x^2)}{x^2}\cdot x^2\ln(x)=\lim_{x\to0} \frac{\tan(x^2)}{x^2}\cdot \lim_{x\to0}x^2\ln(x)=1\cdot0=0$$

therefore

$$\lim_{x\to0^+} x^{\tan(x^2)}=e^0=1$$

Answer 2

Another way, by standard limit $x^x\to 1$ as $x \to 0^+$, since $f(x)=x^{\tan(x^2)}>0$ we have

$$L^2=(f(x))^2=x^{2\tan(x^2)}=\left(x^2\right)^{\tan(x^2)}=\left[\left(x^2\right)^{x^2}\right]^\frac{\tan(x^2)}{x^2} \to 1^1=1 \iff L=1$$

Answer 3

Using a series expansion of $\ln(x)$ about $x=1$:

$$\tan(x^2)\ln(x)= \tan(x^2)\cdot\frac12\left(\dots+4x-3\right)$$

so

$$\lim_{x\to 0^+}\frac12\tan(x^2)(4x-3)=\frac12\tan(0)(4\cdot0-3)=0$$

and

$$\lim_{x\rightarrow0^+}x^{\tan(x^2)}=e^0=1$$