$\lim_{x\to 3}((x+1)(x-3)\tan(\frac{x\pi}{2}))$

Question

Been wondering on how to compute this limit without:

• L'hôpital
• Any Taylor Series

Only with trigonometric identities: $\lim_{x\to 3}((x+1)(x-3)\tan(\frac{x\pi}{2}))$

And yeah I've tried some stuff: $$=\lim_{x\to 3}\frac{(x+1)(x-3)2\tan\left(\frac{xπ}{4}\right)}{1-\tan^2\left(\frac{xπ}{4}\right)}$$ by the double angle identity of tan. Then, I expressed tan as sin and cos and cancelled one $\cos(\frac{x\pi}{4})$ $$=\lim_{x\to 3}\frac{\frac{(x+1)(x-3)2\sin(\frac{xπ}{4})}{\cos(\frac{xπ}{4})}}{\frac{\cos^2(\frac{xπ}{4})-\sin^2(\frac{xπ}{4})}{\cos^2(\frac{xπ}{4})}}$$ $$\lim_{x\to 3}\frac{(x-1)(x-3)2\cos(\frac{x\pi}{4})\sin(\frac{xπ}{4})}{\cos^2(\frac{xπ}{4})-\sin^2(\frac{xπ}{4})}$$ then I used $\cos^2(\alpha)=1-\sin^2(\alpha)$ and $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ and simplified to $$\lim_{x\to 3}\frac{(x+1)(x-3)\sin(\frac{x\pi}{2})}{1-2\sin^2(\frac{x\pi}{4})}$$ but I feel like I'm stuck here.

Answer 1

Hint: As $\lim_{x\to3}(x+1)$ is non-zero finite

$$\lim_{x\to3}(x+1)(x-3)\tan\dfrac{\pi x}2=\lim_{x\to3}(x+1)\cdot\lim_{x\to3}(x-3)\tan\dfrac{\pi x}2$$

Set $x-3=h,$ then $$\tan\dfrac{\pi x}2=\tan\left(\dfrac{\pi(3+h)}2\right)=\tan\left(\dfrac{3\pi}2+\dfrac{\pi h}2\right)=-\cot\dfrac{\pi h}2$$

Use $\lim_{h\to0}\dfrac{\sin h}h=1.$