# Thread: Help with a limit

1. ## Help with a limit

Hi guys

I am not that good with natural log and exponential functions

2. This is good

$\displaystyle \ln(y)=\frac{\sin(x)}{\cos(x)}\ln(\sin(x))=\frac{1 }{\cos(x)}\frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}$

Now the 2nd factor is an indeterminate from so L'hospitials rule applies.

Can you finish from here?

3. could you help me more! I am kinda stuck again

4. Originally Posted by metallica007
could you help me more! I am kinda stuck again
Remember that

$\displaystyle \lim_{x \to a}f(x)g(x)=f(a)g(a)$

$\displaystyle \ln(y)=\frac{\sin(x)}{\cos(x)}\ln(\sin(x))=\frac{1 }{\cos(x)}\frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}$

So we just need to analyze the 2nd factor

$\displaystyle \frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}$

Using L'hosipitals rule we get

$\displaystyle \lim_{x \to 0}\frac{\ln(\sin(x))}{\frac{1}{\sin(x)}}=\lim_{x \to 0} \frac{\frac{\cos(x)}{\sin(x)}}{\frac{-\cos(x)}{\sin^2(x)}}=\lim_{x \to 0}-\sin(x)=0$

This tells us the limit is

$\displaystyle \ln(y)=0 \implies y=1$