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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.

Xetrez
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1 Answers1

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

Bumblebee
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lab bhattacharjee
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  • I've always wondered, doesn't the property $\lim_{x\to a}f(x)g(x)=\lim_{x\to a}f(x)\cdot \lim_{x\to a}g(x)$ only work if the limits as x approaches a of f(x) and of g(x) exist? – Xetrez Jan 23 '21 at 17:07
  • @Xetrez, https://math.stackexchange.com/questions/2674338/proof-of-product-rule-for-limits and https://en.wikibooks.org/wiki/Calculus/Proofs_of_Some_Basic_Limit_Rules – lab bhattacharjee Jan 23 '21 at 17:10
  • yeah, that's the proof, but we don't know here that $\lim_{x\to 3}(x-3)\tan(\frac{\pi x}{2})=L$. In fact, you are assuming so, but i guess we can work with it and instead be unrigorous? – Xetrez Jan 23 '21 at 17:12
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    @Xetrez, Please follow the condition mentioned in the first line of the post. As long as a limit is finite & non-zero, we can safely take it out of the product – lab bhattacharjee Jan 23 '21 at 17:16
  • hmm, okay that's now clear, but how is $\tan(\frac{3\pi}{2}+\alpha)=\cot(-\alpha)$? does this come from the unit circle? because if you use the addition formula it gives one term $\tan(\frac{3\pi}{2})$ which is undefined – Xetrez Jan 23 '21 at 17:48
  • @Xetrez, https://www.math-only-math.com/all-sin-tan-cos-rule.html#:~:text=What%20is%20all%20sin%20tan,the%20fourth%20quadrant%20are%20positive. – lab bhattacharjee Jan 23 '21 at 17:50
  • Let us continue this discussion in chat. – Xetrez Jan 23 '21 at 17:58