Question

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

• A. $\frac{b}{a}$
• B. $\frac{nb}{a}$
• C. $\frac{a}{b}$
• D. $\frac{na}{b}$

Answer 1

The correct option is B $\frac{nb}{a}$

$\underset{x\to a}{lim}\left(\frac{\sqrt{f\left(x\right)}-a}{\sqrt{\varphi \left(x\right)}-b}\right)$

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

$=\underset{x\to a}{lim}\left(\frac{f^{\prime }\left(x\right)}{{\varphi }^{\prime }\left(x\right)}\right)\left(\frac{2b}{2a}\right)$ ...... Using L-Hospital's rule

$=\frac{f^{\prime }\left(a\right)}{{\varphi }^{\prime }\left(a\right)}\left(\frac{b}{a}\right)$

$=\frac{nb}{a}$