Limit of $( n\sin^2(x\pi)-n\sin^2(\sqrt{n}\pi))/(x-\sqrt{n})$ without using L'Hopital 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:
$$ \lim_{x\to \sqrt{n}^+} \frac{n\sin^2(x\pi)-n\sin^2(\sqrt{n}\pi)}{x-\sqrt{n}} =
n\pi\sin(2\pi\sqrt{n})$$
 A: 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 ...
A: Use that:
$$\frac{n\sin^2\pi x-n\sin^2\pi\sqrt{n}}{x-\sqrt{n}}=n\frac{(\sin\pi x-\sin\pi\sqrt{n})(\sin\pi x+\sin\pi\sqrt{n})}{x-\sqrt{n}}=\\
2n(\sin\pi x+\sin\pi\sqrt{n})\cos\left(\frac{\pi\left(x+\sqrt{n}\right)}{2}\right)\frac{\sin\frac{\pi(x-\sqrt{n})}{2}}{x-\sqrt{n}}.$$
A: 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
$$
\lim_{x\to \theta}\frac{\sin(\mp x\pm\theta)}{\mp x\pm\theta}=1 \quad \lim_{x\to \theta}\frac{1-\cos (\mp x\pm\theta)}{\mp x\pm\theta}=0
$$
