I was trying to solve this problem:
Find the angle between the one-sided tangents to the curve at the point x = 0. $$f(x) =\sqrt{1-e^{-x^2}}$$ I obtained the point (0,0) and I have computed the first derivative $$f'(x)=\frac{-xe^{-x^2}}{\sqrt{1-e^{-x^2}}}$$ but when I try to substitute by $0$ I get that the slope is $0$. How can I solve it ?
There are a couple of mistakes:
First, $$f'(x) = \frac{xe^{-x^2}}{\sqrt{1-e^{-x^2}}}. \tag 1$$
Now, it is easy to prove that $$\lim_{x \to 0^{+}}f'(x)=1, \tag 2$$ and $$\lim_{x \to 0^{-}}f'(x)=-1. \tag 3$$ Therefore, the tangents are $y=x$ and $y=-x$, and the angle between them is $\pi/2$.