Science:Math Exam Resources/Courses/MATH100/December 2018/Question 03/Solution 1

We denote the limit by $L$ and next we compute

$\log (L)=\underset{x\to 0^{+}}{\lim }\sin (x)\cdot \log (x)=\underset{x\to 0^{+}}{\lim }\frac{\log (x)}{\frac{1}{\sin (x)}}.$

We see that represented as a quotient above, then the limit for $\log (L)$ is of the form $-\mathrm{\infty }/\mathrm{\infty }$ and so, we apply L'Hôpital's Rule:

$\log (L)=\underset{x\to 0^{+}}{\lim }\frac{\frac{1}{x}}{\frac{-\cos (x)}{{\sin }^2(x)}}=\underset{x\to 0^{+}}{\lim }\frac{{\sin }^2(x)}{-x\cos (x)}.$

This limit is of the form $0/0$, so we apply again L'Hôpital's Rule and obtain:

$\log (L)=\underset{x\to 0^{+}}{\lim }\frac{2\sin (x)\cos (x)}{-\cos (x)-x\cdot (-\sin (x))}=0.$

Since $\log (L)=0$, we conclude that $L=1$.