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If the function \mathrm{f} is differentiable and strictly increasing in a neighbourhood of 0 , then \mathrm{\lim _{x \rightarrow 0} \frac{f\left(x^{5}\right)-f(x)}{f(x)-f(0)}}is equal to

Option: 1

-1


Option: 2

0


Option: 3

1


Option: 4

5/2


Answers (1)

best_answer

\mathrm{\frac{f\left(x^{5}\right)-f(x)}{f(x)-f(0)}}
\mathrm{=\frac{f\left(x^{5}\right)-f(0)-(f(x)-f(0))}{f(x)-f(0)}}
\mathrm{=\frac{f\left(x^{5}\right)-f(0)}{f(x)-f(0)}-1}
\mathrm{=\frac{\frac{f\left(x^{5}\right)-f(0)}{x^{5}-0} \times x^{5}}{\frac{f(x)-f(0)}{x-0} \times x}-1}

So \mathrm{\begin{aligned} \quad \lim _{x \rightarrow 0} & \frac{f\left(x^{5}\right)-f(x)}{f(x)-f(0)}=\lim _{x \rightarrow 0} x^{4} \times \frac{f^{\prime}(0)}{f^{\prime}(0)}-1 \\ & =0-1=-1 .\left(f^{\prime}(0)>0\right)\end{aligned}}

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