My task is to find the equivalent of $$\tan\left (\frac{\pi}{2x+1}\right )$$ in zero. I tried using the formula $\tan(x)\sim x$ in zero, and got $\frac{\pi}{2x+1}$ and then this is $\sim\pi$ in zero. But according to the teacher the right answer is $-2\pi x$
Can you please point me to the right direction?
We have $f(x) = \tan \left(\frac{\pi}{2x+1} \right)$, and $f'(x) = -\frac{2 \pi \sec^2 \pi/(2x+1)}{(x+1)^2}$.
Then to find the first-order approximation, we need to find:
$$f(0) + f'(x) (x-0)$$ $$= 0 + -2\pi x$$