f(x)=x2,ϕ(a)=b2,f'(a)=nϕ'(a) and Φ'(a)≠0 then Limx→a(f(x)−aϕ(x)−b)=

By L- Hospital rule Given limit = limx→a12f(x)f'(x)12ϕ(x).ϕ'(x)=nba

f(x)=x2,ϕ(a)=b2,f'(a)=nϕ'(a) and Φ'(a)≠0 then Limx→a(f(x)−aϕ(x)−b)=

By L- Hospital rule Given limit = limx→a12f(x)f'(x)12ϕ(x).ϕ'(x)=nba

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$f (x) = x^{2}, ϕ (a) = b^{2}, f^{'} (a) = n ϕ' (a)$ and $Φ' (a) \neq 0$ then $\underset{x \to a}{L i m} (\frac{\sqrt{f (x)} - a}{\sqrt{ϕ (x)} - b}) =$

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$\frac{a}{b}$

$\frac{n b}{a}$

$\frac{n a}{b}$

$\frac{b}{a}$

answer is B.

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Detailed Solution

By L- Hospital rule Given limit = $\underset{x \to a}{^{\lim}} \frac{\frac{1}{2 \sqrt{f (x)}} f^{'} (x)}{\frac{1}{2 \sqrt{ϕ (x)}} . ϕ' (x)} = \frac{n b}{a}$

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f(x)=x2,ϕ(a)=b2,f'(a)=nϕ'(a) and Φ'(a)≠0 then Limx→a(f(x)−aϕ(x)−b)=

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$f (x) = x^{2}, ϕ (a) = b^{2}, f^{'} (a) = n ϕ' (a)$ and $Φ' (a) \neq 0$ then $\underset{x \to a}{L i m} (\frac{\sqrt{f (x)} - a}{\sqrt{ϕ (x)} - b}) =$