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

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

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

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

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

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

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

Since $\displaystyle \log(L)=0$, we conclude that $\color {blue}\displaystyle L=1$.