Question

If $f(a) = a^2 , \phi (a) = b^2 $ and $f'(a) = 3 \phi '(a)$ then $Lt_{x \to a} \frac{\sqrt{ f(x)} - a}{ \sqrt{\phi (x)} - b}$ is

• A. $\frac{b^2}{b^2}$
• B. $\frac{b}{a}$
• C. $\frac{2b}{a}$
• D. $\frac{3b}{a}$

Answer 1

The Correct Option is D

$Lt_{x\to0} \frac{\sqrt{f\left(x\right)} -a}{\sqrt{\phi\left(x\right)} -b}$ $ =Lt _{x \to a} \frac{f\left(x\right) -a^{2}}{\phi\left(x\right)-b^{2}} . \frac{\sqrt{\phi\left(x\right)} + b}{\sqrt{f\left(x\right)} +a} $ $= Lt_{x\to a} . \frac{f\left(x\right) -f\left(a\right)}{\phi\left(x\right) -\phi\left(a\right)} . \frac{\sqrt{\phi\left(x\right)} +b}{\sqrt{f\left(x\right)} +a}$ $ = \frac{f'\left(a\right) \left(\sqrt{\phi\left(a\right)} + b\right)}{\phi'\left(a\right)\left(\sqrt{f\left(a\right)} + a\right)} = 3. \frac{b+b}{a+a} =3. \frac{2b}{2a} = \frac{3b}{a} $