Answer by Elias Costa for Limit of $(...
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 $$2\sin \frac{a+b}{2} \cos \frac{a-b}{2} = {\sin(b) +...
View ArticleAnswer by P.. for Limit of $(...
Use that:$$\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...
View ArticleAnswer by Brian M. Scott for Limit of $(...
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...
View ArticleLimit of $( n\sin^2(x\pi)-n\sin^2(\sqrt{n}\pi))/(x-\sqrt{n})$ without using...
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,...
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