6
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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 ...

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – jreing Jan 8 '13 at 14:14
  • $\begingroup$ 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. $\endgroup$ – Michael E2 Jan 8 '13 at 14:15
  • 1
    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Jan 8 '13 at 14:18
  • $\begingroup$ @user1685224 : The point is that it's the limit of a difference quotient, and that's exactly the definition of "derivative". $\endgroup$ – Michael Hardy Jan 8 '13 at 14:40
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks Elias, I figured out how to do it using your hints! Thanks a lot. $\endgroup$ – jreing Jan 8 '13 at 14:32
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.