What is the derivative of $f(x) = e^x + x^3 cos(\frac{1}{x})$, where f(0) = 1, at x = 0?

Question

Intuitively, I would think the derivative of the function at x = 0 would be 0, since f(0) = 1, but I am not sure if there is supposed to be a more rigorous way of doing this. Taking the same approach from this post: Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$ I find that

$\lim_{h\to0} \frac{f(h) - f(0)}{h} = \lim_{h\to0} \frac{e^h + h^3 cos(\frac{1}{h}) - 1}{h}$

This seems to imply that the derivative doesn't exist. Is this the case or am I on the wrong track? Any help would be greatly appreciated.

Answer 1

You're almost there, except that all your limits should be as $h\to 0$ (not $h\to\infty$). This being fixed, rewrite, for $h\neq 0$, $$ \frac{e^h + h^3 \cos \frac{1}{h} - 1}{h} = \frac{e^h -1 }{h} + h^2 \cos \frac{1}{h} $$ Now, $$ \lim_{h\to 0} h^2 \cos \frac{1}{h} = 0 $$ since $\cos$ is bounded (use the Squeeze theorem). And $$ \lim_{h\to 0} \frac{e^h -1 }{h} = \exp'(0) = e^0 = 1 $$ recognizing the definition of the derivative of $\exp$ at $0$.

That leads to $$ \lim_{h\to 0}\frac{e^h + h^3 \cos \frac{1}{h} - 1}{h} = \boxed{1}\,. $$

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An interesting question is why did you think that $f(0)=1$ would imply $f'(0)=0$? There is no reason for that to hold in general...

Answer 2

It should be $h\rightarrow 0$, so

\begin{align*} \lim_{h\rightarrow 0}\dfrac{f(h)-f(0)}{h}&=\lim_{h\rightarrow 0}\dfrac{e^{h}+h^{3}\cos(1/h)-1}{h}\\ &=\lim_{h\rightarrow 0}\dfrac{1+h+h^{2}/2!+\cdots+h^{3}\cos(1/h)-1}{h}\\ &=\lim_{h\rightarrow 0}\dfrac{h+h^{2}/2+\cdots+h^{3}\cos(1/h)}{h}\\ &=\lim_{h\rightarrow 0}\left(1+h/2+\cdots+h^{2}\cos(1/h)\right)\\ &=1. \end{align*}

Answer 3

$$\lim_{h\to0} \frac{f(h) - f(0)}{h} = \lim_{h\to0} \frac{e^h + h^3 cos(\frac{1}{h}) - 1}{h}=$$

$$\lim_{h\to0} \frac{e^h - 1}{h} + \lim_{h\to0} \frac { h^3 cos(\frac{1}{h})}{h}=1+0=1 $$