# Proving limit of log-trig function $\lim \limits_{x \downarrow 0} \big(1+x^2 \log x\big)^{\csc(x)}$

I am trying to evaluate the following limit (if it exists): $$\lim \limits_{x \downarrow 0} \big(1+x^2 \log x\big)^{\csc(x)}$$ where $\log x$ is the natural logarithm. Intuitively, I think the $x^2$ should dominate over the logarithmic term in the limit. I also know that for $x$ small enough $\sin x\approx x$, so $\csc x \approx 1/x$. Preliminary computations suggest the limit is 1, but I am not sure how to approach this formally. If the limit indeed exists, are there any tricks at my disposal?

Take the $\log$ of the limit so that we're evaluating

$$\lim \frac{\log({1+x^2\log x})}{\sin x} = \lim \frac{\log(1+x^2\log x)}{x} = \lim\ (x\log x) \left(\frac{\log(1+x^2\log x)}{x^2 \log x}\right)$$

Firstly, note that $\frac{\log(1+t)}{t} \to 1$, in our case $t = x^2\log x$ which tends to $0$ as $x \to 0$ (this is a corollary of the next fact).

Further, note $x \log x \to 0$. This can be seen by letting $x = e^{-u}$ so that we're evaluating $\frac{-u}{e^u}$, $u \to \infty$,which is clear by, for instance, considering the taylor expansion of $\exp$.

This implies the above line of equalities is equal to $0$ so $\log \lim = 0$. This implies our limit is $1$.

$$\lim \limits_{x \to 0} \big(1+x^2 \log x\big)^{\csc(x)}$$ Using second remarkable limit: $$\lim \limits_{x \to 0} \big(1+x^2 \log x\big)^{\frac{x^2\log(x)}{x^2\log(x)}\csc(x)}=\lim \limits_{x \to 0} e^{x^2\log(x)\csc(x)}$$ Then, consider limit $\lim\limits_{x \to 0} x^2\log(x)\csc(x)=\lim\limits_{x \to 0} \frac{x^2\log(x)}{\sin(x)}$

Using first remarkable limit: $$\lim\limits_{x \to 0}\frac{x^2\log(x)}{\sin(x)}=\lim\limits_{x \to 0} x\log(x)$$ And L'Hospital's Rule: $$\lim\limits_{x \to 0} x\log(x)=\lim\limits_{x \to 0} \frac{\log(x)}{\frac{1}{x}} =\lim\limits_{x \to 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim\limits_{x \to 0} (-x)=0$$ Which leads to: $$\lim \limits_{x \to 0} \big(1+x^2 \log x\big)^{\csc(x)}=\lim \limits_{x \to 0} e^{x^2\log(x)\csc(x)}=e^0=1$$