Is this the correct way of solving Calculus infinity limit?

Question

in this question, can I just do direct substitution? $$\lim_{x\to 0^+}\ln(\sin(x))$$

Thanks!

Answer 1

You can’t substitute $x=0$, since $\ln 0$ is undefined, but you can use the fact that $\sin x\to 0^+$ as $x\to 0^+$ to say that

$$\lim_{x\to 0^+}\ln\sin x=\lim_{x\to 0^+}\ln x\;.$$

That’s a limit that you should know: $$\lim_{x\to 0^+}\ln x=-\infty\;.$$

Answer 2

That works when the function is continuous at the limit point. As it stands, the function is not (since it isn't defined there), so I wouldn't look at it that way. However you can substitute $u= \sin x$. Then as $x\to 0^+, u\to 0^+$, so

$$\lim_{x\to 0^+}\ln(\sin x)=\lim_{u\to 0^+}\ln u$$

The limit diverges to $-\infty$. It seems like you're trying to say that $\ln(0)=-\infty$, which is incorrect.

Answer 3

Try

$\ln( \lim_{x\to 0^+} \sin(x) )$

So as $x \to 0^+$, $\ln(\sin(x) ) \to -\infty$