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I was asked to prove this , without using L'Hopital... tried out some trig. identities with no big use $(\sin(\alpha)-\sin(\beta))(\sin(\alpha)+\sin(\beta))=\sin^2(\alpha)-\sin^2(\beta)$ for example, and from there to the $\sin(\alpha)-\sin(\beta)$ identity... but with no real success. And tried also multiplying num.and denum. by the conjugate.

the question is: Prove (without using L'Hopital) that: $$ \lim_{x\to \sqrt{n}^+} \frac{n\sin^2(x\pi)-n\sin^2(\sqrt{n}\pi)}{x-\sqrt{n}} = n\pi\sin(2\pi\sqrt{n})$$

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3 Answers 3

Note that the result is equivalent to

$$\lim_{x\to\sqrt{n}^+}\frac{\sin^2\pi x-\sin^2\pi\sqrt{n}}{x-\sqrt{n}}=\pi\sin 2\pi\sqrt{n}\;.\tag{1}$$

HINT for $(1)$: Let $f(x)=\sin^2\pi x$. What is the definition of $f\,'(\sqrt n)$? (And you may want a double angle formula as well.)

Added: Perhaps I should have emphasized the word definition in the hint. The lefthand side of $(1)$ is the limit of a difference quotient ...

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Thanks for the quick responese. I used the hint and got to this point: $f'(\sqrt{n}) = \sin(2\pi\sqrt{n})$ . I don't understand why I am allowed to use the derivative for this question. and also how to get from this to getting rid of the denum. and acheiving that $\pi$ before the $\sin$ in the answer... –  user1685224 Jan 8 '13 at 14:14
    
Funny problem. You could also replace $\sqrt{n}$ by $c$. The $n$ and $\sqrt{n}$ are sort of red herrings, perhaps there just to obfuscate the problem a bit. –  Michael E2 Jan 8 '13 at 14:15
1  
@user1685224: No, the derivative is $2\pi\sin\pi\sqrt{n}\cos\pi\sqrt{n}=\pi\sin 2\pi\sqrt{n}$, which is exactly what you need. I don’t know whether you’re allowed to use the derivative, but it’s definitely not a use of l’Hospital’s rule. –  Brian M. Scott Jan 8 '13 at 14:18
    
@user1685224 : The point is that it's the limit of a difference quotient, and that's exactly the definition of "derivative". –  Michael Hardy Jan 8 '13 at 14:40

We have $$ \sin(x\pi)^2-\sin( \sqrt{n}\pi)^2= (\sin(x\pi)+\sin( \sqrt{n}\pi))\cdot (\sin(x\pi)-\sin( \sqrt{n}\pi)) $$ Use the trigonometric formulas $$2\sin \frac{a+b}{2} \cos \frac{a-b}{2} = {\sin(b) + \sin(a) } \\ 2\sin \frac{a-b}{2} \cos \frac{a+b}{2} = {\sin(b) - \sin(a) } $$ and fundamental trigonometric limits $$ \lim_{x\to \theta}\frac{\sin(\mp x\pm\theta)}{\mp x\pm\theta}=1 \quad \lim_{x\to \theta}\frac{1-\cos (\mp x\pm\theta)}{\mp x\pm\theta}=0 $$

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Thanks Elias, I figured out how to do it using your hints! Thanks a lot. –  user1685224 Jan 8 '13 at 14:32

Use that: $$\frac{n\sin^2\pi x-n\sin^2\pi\sqrt{n}}{x-\sqrt{n}}=n\frac{(\sin\pi x-\sin\pi\sqrt{n})(\sin\pi x+\sin\pi\sqrt{n})}{x-\sqrt{n}}=\\ 2n(\sin\pi x+\sin\pi\sqrt{n})\cos\left(\frac{\pi\left(x+\sqrt{n}\right)}{2}\right)\frac{\sin\frac{\pi(x-\sqrt{n})}{2}}{x-\sqrt{n}}.$$

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