Limit to differentiate

Calculus Level 2

Let $f$ be a real-valued differentiable function on the real line $\mathbb{R}$ such that $\lim_{x\to 0} \dfrac{f(x)}{x^2}$ exists, and is finite . Then the value of $f'(0)$ is

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1 solution

Ron Gallagher
Dec 2, 2020

Since f(x)/x^2 approaches a finite limit as x->0, this means that f(x) ->0 as x->0. Since f is differentiable (and, hence continuous) at x = 0, this implies that f(0) = 0. We then have, as x->0:

lim(f(x) / x^2) = lim((1/x) * (f(x)/x)) = lim((1/x) * (f(x) - f(0))/(x-0)).

Since this limit exists and is finite, this implies:

(f(x) - f(0)) / (x-0) approaches 0 as x->0. By definition, this means that f'(0) = 0.

Limit to differentiate

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