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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?

FFjet
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Loukil-M
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    This is not clear. What do you mean by "in zero"? – Klangen Sep 05 '19 at 11:32
  • @Klangen I'm not sure what's called in english. According to the definition $f\sim_a g\Leftrightarrow\lim_{x\to a}\frac{f(x)}{g(x)}=1$, i'm asking here when a = 0. – Loukil-M Sep 05 '19 at 11:39
  • It sounds like you are trying to find a linear approximation around zero. – Toby Mak Sep 05 '19 at 11:40
  • If your task is to find a function $g$ such that $\lim_{x\to a}\frac{f(x)}{g(x)}=1$ then there would be many choices for $g$; I suppose there is a further constraint, namely, $g$ should be of the form $g(x)=mx+b$? Even then the answer is unique only if $f(a)=0.$ But a "linear approximation" would be unique if $f$ is differentiable at $a.$ – David K Sep 05 '19 at 11:52

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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:

$$f(0) + f'(x) (x-0)$$ $$= 0 + -2\pi x$$

Toby Mak
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  • Thanks! It's the first time i hear about linear approximation. – Loukil-M Sep 05 '19 at 11:51
  • This comes from the general form for a Taylor series around zero: $f(0) + f'(x)(x-0) + \frac{1}{2!}f''(x) \cdot (x-0)^2 + \frac{1}{3!} f^{(3)} (x) \cdot (x-0)^3 + \cdots$. and making an approximation by discarding the higher-order terms. – Toby Mak Sep 05 '19 at 11:52