Question

$\underset{\alpha \to 0}{lim}\frac{{\int }_0^{\alpha }xdx}{\alpha \sin \alpha }$ is equal to

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $\frac{1}{2}$

The given limit is of the form $\frac{0}{0}$

Hence, using L'Hopitals' Rule:-

$\underset{\alpha \to 0}{lim}\frac{{\int }_0^{\alpha }xdx}{\alpha \sin \alpha }$

According Leibniz integral rule-

$\frac{d}{dx}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x\right)dx$

$=f\left(b\right(x\left)\right)\frac{d}{dx}b\left(x\right)-f\left(a\right(x\left)\right)\frac{d}{dx}a\left(x\right)$

$\underset{\alpha \to 0}{lim}\frac{\frac{d}{d\alpha }\left({\int }_0^{\alpha }xdx\right)}{\frac{d}{d\alpha }\left(\alpha \sin \alpha \right)}$

$=\underset{\alpha \to 0}{lim}\frac{\alpha }{\sin \alpha +\alpha \cos \alpha }$

$=\underset{\alpha \to 0}{lim}\frac{1}{\frac{\sin \alpha }{\alpha }+\cos \alpha }$

$=\frac{1}{1+1}=\frac{1}{2}$

Hence, answer is option-(B).