Question

Let $f\left(x\right)$ be a strictly increasing and differentiable function, then
$\underset{x\to 0}{lim}\frac{f\left(x^2\right)-f\left(x\right)}{f\left(x\right)-f\left(0\right)}$, equals

• A. $2$
• B. $1$
• C. $-1/2$
• D. $-1$

Answer 1

The correct option is D $-1$

$\underset{x\to 0}{lim}\frac{f\left(x^2\right)-f\left(x\right)}{f\left(x\right)-f\left(0\right)}=\underset{x\to 0}{lim}\frac{f\left(x^2\right)+f\left(0\right)-f\left(0\right)-f\left(x\right)}{x-0}\times \frac{x-0}{f\left(x\right)-f\left(0\right)}$$=\underset{x\to 0}{lim}(x.\frac{f\left(x^2\right)-f\left(0\right)}{x^2-0}-\frac{f\left(x\right)-f\left(0\right)}{x-0}).\frac{x-0}{f\left(x\right)-f\left(0\right)}$

$=\left(0.f^{\prime }\right(0)-f^{\prime }(0\left)\right).\frac{1}{f^{\prime }\left(0\right)}=-1$

(provided $f^{\prime }\left(0\right)\ne 0)$