Evaluate the limit: $\lim_{x\to \frac {\pi}{2}} \frac {\cos x}{\log (x-\frac {\pi}{2} +1)}$

Question

Evaluate the limit: $$\lim_{x\to \frac {\pi}{2}} \frac {\cos x}{\log (x-\frac {\pi}{2} +1)}$$

... The given function takes indeterminate form $(\frac {0}{0})$ when $x=\frac {\pi}{2}$. Let $x-\frac {\pi}{2}=y$, $x=y+\frac {\pi}{2}$. As $x\to \frac {\pi}{2}$; $y\to 0$. Now, $$=\lim_{y\to 0} \frac {\cos (y+\frac {\pi}{2})}{\log (y+1)}$$ $$=\lim_{y\to 0} \frac {-\sin \frac {\pi}{2}}{\log (y+1)}$$

Answer 1

With your substitution we obtain: $$\lim_{y\rightarrow0}\frac{-\sin{y}}{\ln(1+y)}=\lim_{y\rightarrow0}\frac{-\sin{y}}{y}\lim_{y\rightarrow0}\frac{y}{\ln(1+y)}=-1\cdot1=-1$$

Answer 2

There is no need to make any substitution in this problem since one can apply L'Hopital's Rule immediately once one has verified that the conditions are met. Thus

\begin{eqnarray} \lim_{x\to \frac {\pi}{2}} \frac {\cos x}{\log (x-\frac {\pi}{2} +1)}&=&\lim_{x\to \frac {\pi}{2}}(-\sin x)(x-\frac {\pi}{2} +1)\\ &=&(-1)(1)\\ &=&-1 \end{eqnarray}

Note: This is assuming that by $\log$ you mean natural logarithm. If by $\log$ you mean common logarithm with base $10$ then the limit will be $-\ln(10)$.