Limit of $( n\sin^2(x\pi)-n\sin^2(\sqrt{n}\pi))/(x-\sqrt{n})$ without using L'Hopital

Question

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: $$ \lim_{x\to \sqrt{n}^+} \frac{n\sin^2(x\pi)-n\sin^2(\sqrt{n}\pi)}{x-\sqrt{n}} = n\pi\sin(2\pi\sqrt{n})$$

Answer 1

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 ...

Answer 2

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

Answer 3

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 $$ \lim_{x\to \theta}\frac{\sin(\mp x\pm\theta)}{\mp x\pm\theta}=1 \quad \lim_{x\to \theta}\frac{1-\cos (\mp x\pm\theta)}{\mp x\pm\theta}=0 $$