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

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.

## 3 Answers

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

therefore

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

• Thank you! Didn't think of multiplying by (x²/x²), that's very clever! Jun 5 at 19:00

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

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

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

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