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Use L'Hopital to calculate

$$\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$$

Right now this yields $\frac{0}{0}$ so let's go ahead and use L'Hopital:

$$\lim_{x\to +\infty}\frac{\frac{2}{x^3}}{2\cdot \sin\left(\frac{2}{x}\right)\cdot\cos \left(\frac{2}{x}\right)\cdot\left(\frac{-2}{x^2}\right)}$$

This just won't do. Perhaps we should flip the numerator with the denominator instead:

$$-\frac{\csc^2\left(\frac{2}{x}\right)}{x^{2}}$$

This yields -$\frac{\infty}{\infty}$, so we can go ahead and apply L'Hopital:

$$-\frac{2\cdot-\csc\left(\frac{2}{x}\right)\cdot\cot\left(\frac{2}{x}\right)\cdot\frac{-2}{x^2}}{2x}$$

If I evaluate this I will get $\frac{0}{\infty}$

I have the feeling I'm not supposed to keep going this path, and there's a simpler solution (using L'Hopital). What can I do to solve this?

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Perhaps

$$\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}=-\frac14\lim_{x\to +\infty}\left(\frac{\frac{2}{x}}{\sin\left(\frac{2}{x}\right)}\right)^2=-\frac14$$

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  • $\begingroup$ I can see and understand why this works. However, it's not clear to me what triggered your idea of dividing $\frac{-1}{x^2}$ by $\frac{-1}{4}$. What made you see that? $\endgroup$ – Zol Tun Kul Nov 10 '15 at 2:24
  • $\begingroup$ After countless exercises, the fact that $\lim_{x\rightarrow0} \frac{\sin(x)}x=1$ has been engraved in my mind. I saw a $\sin^2$ and an $x^2$ and I thought: "This looks almost as that good ol' sine limit, what am I missing to be able to use it?" $\endgroup$ – EA304GT Nov 10 '15 at 2:33
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Use the substitution $u=\frac{2}{x}$

Then the limit becomes

\begin{align} \lim_{u\to 0} \frac{-\frac{1}{4}u^2}{\sin^2{u}}&=\lim_{u\to 0}\frac{-\frac{1}{2}u}{2\sin u\cos u}\\ \\ &=-\frac{1}{4}\lim_{u\to 0}\frac{u}{\sin u\cos u}\\ \\ &=-\frac{1}{4}\lim_{u\to 0}\frac{1}{\cos^2 u-\sin^2 u}\\ \\ &=-\frac{1}{4} \end{align}

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  • $\begingroup$ Why did you pick $\frac{2}{x}$ in particular? I mean, I can see why your solution works, but I'm having a hard time understanding your reasoning behind picking $\frac{2}{x}$ and not something else. Is it because of the sine function? $\endgroup$ – Zol Tun Kul Nov 10 '15 at 3:24
  • $\begingroup$ Only because of a personal issue with sine functions. As you can see from another user, $1/x$ works fine. But for me, my first instinct is to so something to simplify trig functions since I find them more annoying handle. $\endgroup$ – Elliot G Nov 10 '15 at 3:29
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Let $t=\frac{1}{x},$ then \begin{equation*} \lim_{x\rightarrow +\infty }\frac{-\frac{1}{x^{2}}}{\sin ^{2}\left( \frac{2}{% x}\right) }=\lim_{t\rightarrow 0^{+}}\frac{-t^{2}}{\sin ^{2}\left( 2t\right) }=\frac{-1}{4}\lim_{t\rightarrow 0^{+}}\left( \frac{(2t)}{\sin \left( 2t\right) }\right) ^{2}=\frac{-1}{4}\left( \lim_{u\rightarrow 0^{+}}\frac{u}{% \sin u}\right) ^{2}=-\frac{1}{4}\cdot 1=-\frac{1}{4}. \end{equation*} One can use L'Hospital's rule to verify that \begin{equation*} \lim_{u\rightarrow 0^{+}}\frac{u}{\sin u}=\lim_{u\rightarrow 0^{+}}\frac{1}{% \cos u}=\frac{1}{\cos 0}=1. \end{equation*}

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$$\lim_{x\to\infty}\frac{-\frac{1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}=$$ $$-\left(\lim_{x\to\infty}\frac{\frac{1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}\right)=$$ $$-\left(\lim_{x\to\infty}\frac{\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^2}\right)}{\frac{\text{d}}{\text{d}x}\left(\sin^2\left(\frac{2}{x}\right)\right)}\right)=$$ $$-\left(\lim_{x\to\infty}\frac{-\frac{2}{x^3}}{-\frac{4\sin\left(\frac{2}{x}\right)\cos\left(\frac{2}{x}\right)}{x^2}}\right)=$$ $$-\left(\lim_{x\to\infty}\frac{\csc\left(\frac{2}{x}\right)\sec\left(\frac{2}{x}\right)}{2x}\right)=$$ $$-\left(\lim_{x\to\infty}\frac{1}{2x\cos\left(\frac{2}{x}\right)\sin\left(\frac{2}{x}\right)}\right)=$$ $$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)\left(\lim_{x\to\infty}\frac{1}{\cos\left(\frac{2}{x}\right)}\right)=$$ $$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)\left(\frac{1}{\lim_{x\to\infty}\cos\left(\frac{2}{x}\right)}\right)=$$ $$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)\left(\frac{1}{\cos\left(\lim_{x\to\infty}\frac{2}{x}\right)}\right)=$$ $$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)=-\frac{1}{2}\cdot\frac{1}{2}=-\frac{1}{4}$$

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