in this question, can I just do direct substitution? $\lim_{x\to 0^+}\ln(\sin(x))$
Thanks!
in this question, can I just do direct substitution? $\lim_{x\to 0^+}\ln(\sin(x))$
Thanks!
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: $\lim_{x\to 0^+}\ln x=-\infty\;.$
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.
Try
$\ln( \lim_{x\to 0^+} \sin(x) )$
So as $x \to 0^+$, $\ln(\sin(x) ) \to -\infty$