Indeterminate forms and l'Hospital rule

Question

The graph of a function and its tangent line at $0$ are shown in image. What is the value of $\lim_{x \to 0} \frac{f(x)}{e^x-1}$?

Answer 1

It seems to go like this: from the figure we conclude $\lim_{x\to 0}f(x) = 0$ so we can apply L'hospital. Now, at $x=0$ the derivative of $f$ is the same as the derivative of $y = x$, so $f'(x) = 1$. Thus $\lim_{x\to 0} \frac{f(x)}{e^{x}-1} = \lim_{x\to 0} \frac{1}{e^{x}} = 1$.