Let's do some calculus! (4)

The answer to the unedited question by taking the two-sided limit should be that the limit does not exist.

Checking the left-hand and the right-hand limits, the left-hand limit does not exist for $\lim _{ x\rightarrow 0 }{ \ln { { \left( \sin { x } \right) } } }$ while the right-hand limit is ${ -\infty }$ . But I would still clarify this with all of you first.

However, the answer will be zero if $x\rightarrow { 0 }^{ + }$ . For the sake of clarification, it means "x approaches zero from the right".

For the solution, let us rewrite the original limit into the indeterminate form in which we can work with L'hopital's Rule .

$L=\lim _{ x\rightarrow { 0 }^{ + } }{ \ln { { \left( \sin { x } \right) }^{ \tan { x } } } }$

$=\lim _{ x\rightarrow { 0 }^{ + } }{ \tan { x } \ln { \left( \sin { x } \right) } }$

$=\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \ln { \left( \sin { x } \right) } }{ \frac { 1 }{ \tan { x } } } }$

$=\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \ln { \left( \sin { x } \right) } }{ \cot { x } } }$

As $x\rightarrow { 0 }^{ + }$ , $\ln { \left( \sin { x } \right) \rightarrow -\infty }$ and $\cot { x } \rightarrow \infty$ . Hence, we have the indeterminate form of ${ \infty }/{ \infty }$ .

Applying L'hopital's Rule, differentiating the numerator and denominator individually, we get the following:

$=\lim _{ x\rightarrow { 0 }^{ + } }{ \frac { \cot { x } }{ -\csc { x } \cot { x } } }$

$=-\lim _{ x\rightarrow { 0 }^{ + } }{ \sin { x } }$

$=0$