While I was tutoring a differential calculus course for engineering undergrads, I encountered the following limit and, being sincere, I don't really know how to solve it without using L'Hôpital's Rule:

$$\lim_{x\to 0^+} \frac{\sqrt{\log^2(x)-\sin(x)}}{\log(x) + e^x}$$

Any tips? I've tried and failed to convert it to a normal, known limit. Wolfram says it equals $-1$.

Intuitively we want to ignore the $\sin(x)$ and $e^x$. So we evaluate

$$\frac{\log(x)+e^x}{\log(x)}=1+\frac{e^x}{\log(x)} \rightarrow 1$$

and

$$\frac{\log^2(x)+\sin(x)}{\log^2(x)}=1+\frac{\sin(x)}{\log^2(x)}\rightarrow 1$$

Note that for $x < 1$, $\sqrt{\log^2(x)}=-\log(x)$ and we have our limit.

A start: Imagine $x$ positive and closeto $1$. Then $\lo x$ is "large negative."

Divide top and bottom by $|\log x|$, and look.

Note that when we divide the top by $|\log x|$ we get $$\sqrt{1-\frac{\sin x}{\log^2 x}}.$$