Question

If $\underset{x\to 0}{lim}\frac{\underset{0}{\overset{x}{\int }}\frac{t^2}{\sqrt{a+t}}dt}{bx-\sin x}=1,$ then

• A. $a\in R$
• B. $a=4$
• C. $b=1$
• D. $a=0,b=0$

Answer 1

Answer: (B), (C)

$b=1$

$\underset{x\to 0}{lim}\frac{\underset{0}{\overset{x}{\int }}\frac{t^2}{\sqrt{a+t}}dt}{bx-\sin x}=1$
Since limit is of $\left(\frac{0}{0}\right)$form, therefore by L' hospital's rule, we have
$\underset{x\to 0}{lim}\frac{x^2}{\sqrt{a+x}(b-\cos x)}=1$
Numerator $\to 0,$ while denominator $\to \sqrt{a}(b-1)$ as $x\to 0$
But limit exists and is equal to $1$
$\therefore b-1=0\Rightarrow b=1$
$a\ne 0,$ otherwise limit will not exist.

$\therefore \underset{x\to 0}{lim}\frac{x^2}{\sqrt{a+x}(1-\cos x)}=1$
$\Rightarrow \underset{x\to 0}{lim}\frac{1}{\sqrt{a+x}\left(\frac{1-\cos x}{x^2}\right)}=1$
$\Rightarrow \frac{2}{\sqrt{a}}=1\Rightarrow a=4$