Question

If $f\left(a\right)=a^2,{\varphi }^{\prime }\left(a\right)=b^2$ and f'(a) = $3{\varphi }^{\prime }\left(a\right)$
then $\underset{x\to 0}{lim}\frac{\sqrt{f\left(x\right)}-a}{\sqrt{\varphi \left(x\right)}-b}$ is-

• A. $b^2/a^2$
• B. b/a
• C. 2b/a
• D. None of these

Answer 1

The correct option is D None of these

$f\left(a\right)=a^2{\varphi }^{\prime }\left(a\right)=b^2$
$f^{\prime }\left(a\right)=3{\varphi }^{\prime }\left(a\right)$
Let, $L=\underset{x\to 0}{lim}\frac{\sqrt{f\left(x\right)}-a}{\sqrt{\varphi \left(x\right)-b}},\frac{0}{0}$ form.
So applying L' Hospital's rule,
$L=\underset{x\to 0}{lim}\frac{f^{\prime }\left(x\right)/\sqrt{f\left(x\right)}}{{\varphi }^{\prime }\left(x\right)/\sqrt{\varphi \left(x\right)}}=\frac{f^{\prime }\left(a\right)/\sqrt{f\left(a\right)}}{{\varphi }^{\prime }\left(a\right)/\sqrt{\varphi \left(a\right)}}=\frac{3b}{a}$