# How to calculate the undefined limit of this trig function?

I have to simplify this function : $$f(x) = (x-2)\tan\left(\frac {\pi}{x}\right)$$ In order to calculate $\lim\limits_{x\to2}f(x)$ since $\lim\limits_{x\to2}(x-2) = 0$ and $\lim\limits_{x\to2}\tan\left(\frac {\pi}{x}\right) = \infty$ and their multiplication is undefined. I have tried to replace $x$ with another $X$ so i can apply this rule : $$\lim_{X\to0}\frac{\tan X}{X} = 1$$ but in vain. Some hints would be appreciated thanks! ( excuse my bad english it's not my native language )

## 2 Answers

$$(x-2)\tan\dfrac\pi x=\sin\dfrac\pi x\cdot\frac{x-2}{\cos\dfrac\pi x}$$

$$\text{Now as }\cos\dfrac\pi x=\sin\left(\frac\pi2-\frac\pi x\right)=\sin\frac{\pi(x-2)}{2x},$$

$$\implies(x-2)\tan\dfrac\pi x=\sin\dfrac\pi x\cdot\frac{\dfrac{\pi(x-2)}{2x}}{\sin\dfrac{\pi(x-2)}{2x}}\cdot\frac{2x}\pi$$

Now set $\displaystyle \dfrac{\pi(x-2)}{2x}=y$ to use $\displaystyle\lim_{y\to0}\frac{\sin y}y=1$

• Thanks a lot sir i was on the right track! I got stuck in the last line – Cetos Feb 25 '14 at 17:03
• @Cetos, hope things are clear, now. – lab bhattacharjee Feb 25 '14 at 17:28

Notice that the desired limit is also equal to

$$\lim_{x\to2}\frac{x-2}{\cos\left(\frac \pi x\right)}$$ since $\sin(\pi/2)=1$.

Moreover we have $$\lim_{x\to2}\frac{\cos\left(\frac \pi x\right)}{x-2}=\lim_{x\to2}\frac{\cos\left(\frac \pi x\right)-\cos\left(\frac \pi 2\right)}{x-2}=\left(\cos\left(\frac \pi x\right)\right)'\bigg|_{x=2}=\frac{\pi}{2^2}\sin\left(\frac \pi 2\right)=\frac\pi4$$ hence the desired limit is $\frac4\pi$.

• Excellent explanation! – Namaste Feb 26 '14 at 14:08