How can I find this limit? $$\lim_{x\to0}\left(100\csc^2(x)-\csc^2\left(\frac{x}{10}\right)\right)$$ Please help me to find this limit.
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With these kinds of problems, the first thing I'll usually do is try breaking down CSC and any other trig functions into other trig functions that are easier to deal with. Which usually helps at leas somewhat. – scott_fakename Nov 03 '13 at 07:46
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Try Maclaurin series – user85798 Nov 03 '13 at 08:35
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Hint: write $\csc x=\frac{1}{\sin x}$, make the fractions have the same denominator, multiply and divide with $x^4$, factor the difference of squares, and use L'Hôpitals rule.
detnvvp
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Note that when $x\to0$, $\sin(x)\sim x-\frac{x^3}{6}$ $$\begin{align}\lim_{x\to0}100\csc^2(x)-\csc^2(\frac{x}{10})&=\lim_{x\to0}\frac{100\sin^2(\frac{x}{10})-\sin^2(x)}{\sin^2(x)\sin^2(\frac{x}{10})}\\&=\lim_{x\to0}\left(10\sin(\frac{x}{10})-\sin(x)\right)\times\lim_{x\to0}\frac{10\sin(\frac{x}{10})+\sin(x)}{\sin^2(x)\sin^2(\frac{x}{10})}\\&\sim_0\left(10\left(\frac{x}{10}-\frac{\frac{x^3}{1000}}{6}\right)-x+\frac{x^3}{6}\right)\times\left(\frac{2x}{\frac{x^4}{100}}\right)=33\end{align}$$
user91500
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I apologise for doing this again, but how did you compute the limit with the quotient to be $2x/(x^4/100)$? – Maths Matador Sep 07 '18 at 10:43
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