1
$\begingroup$

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 $-1$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$\sqrt{1-\frac{\sin x}{\log^2 x}}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.