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Mathematics Question on limits and derivatives

If f(a)=a2,ϕ(a)=b2f(a) = a^2 , \phi (a) = b^2 and f′(a)=3ϕ′(a)f'(a) = 3 \phi '(a) then Ltx→af(x)−aϕ(x)−bLt_{x \to a} \frac{\sqrt{ f(x)} - a}{ \sqrt{\phi (x)} - b} is

A

b2b2\frac{b^2}{b^2}

B

ba\frac{b}{a}

C

2ba\frac{2b}{a}

D

3ba\frac{3b}{a}

Answer

3ba\frac{3b}{a}

Explanation

Solution

Ltx→0f(x)−aϕ(x)−bLt_{x\to0} \frac{\sqrt{f\left(x\right)} -a}{\sqrt{\phi\left(x\right)} -b} =Ltx→af(x)−a2ϕ(x)−b2.ϕ(x)+bf(x)+a=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} =Ltx→a.f(x)−f(a)ϕ(x)−ϕ(a).ϕ(x)+bf(x)+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} =f′(a)(ϕ(a)+b)ϕ′(a)(f(a)+a)=3.b+ba+a=3.2b2a=3ba= \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}