Hi guys
Could please help me with this limit:
I am not that good with natural log and exponential functions
Thanks in advance
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
$\displaystyle \ln(y)=0 \implies y=1$