Question

If $f\left(a\right)=a^2,\varphi \left(a\right)=b^2$ and $f^{\prime }\left(a\right)=3{\varphi }^{\prime }\left(a\right)$ 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

$\underset{x\to a}{lim}\frac{\sqrt{f\left(x\right)}-a}{\sqrt{\varphi \left(x\right)}-b}=\underset{x\to a}{lim}\frac{f\left(x\right)-a^2}{\varphi \left(x\right)-b^2}\times \frac{\sqrt{\varphi \left(x\right)}+b}{\sqrt{f\left(x\right)}+a}$
$=\underset{x\to a}{lim}\frac{f\left(x\right)-f\left(a\right)}{\varphi \left(x\right)-\varphi \left(a\right)}\times \frac{\sqrt{\varphi \left(x\right)}+b}{\sqrt{f\left(x\right)}+a}$
$=\frac{f^{\prime }\left(a\right)}{{\varphi }^{\prime }\left(a\right)}\times \frac{\sqrt{\varphi \left(a\right)}+b}{\sqrt{f\left(a\right)}+a}=\frac{3b}{a}$