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Use l'Hopital's rule to find the following limit

$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)$$

I tried differentiation but did not get the right answer. I know I have to put it into a single fraction.

user183782
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  • You need to apply l'Hopital's rule four times. Remember, don't differentiate with the quotient rule; differentiate the numerator and denominator separately. – Empy2 Oct 23 '14 at 11:44
  • Only twice, not four times. – Galc127 Oct 23 '14 at 11:48

3 Answers3

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Putting into a single fraction $$ \csc^2(x)-\frac{1}{x^2} = \frac{1}{\sin^2(x)}-\frac{1}{x^2}=\frac{x^2-sin^2(x)}{x^2sin^2(x)} $$ Try differentiation now. You will get that the limit is $\frac{1}{3}$ as the answer.

Barbosa
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$$\lim_{x\to 0}\left((\csc x)^2-\frac{1}{x^2}\right)=\lim_{x\to 0}\left(\frac{\frac{x^2}{sin^2(x)}-1}{x^2}\right)$$

We get the form $0/0$, so applying L'Hopital for the first time we get $$\displaystyle \lim_{x\to0}\left((\csc x)^2-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\large \frac{\frac{2x \sin^2(x)-2x^2 \sin(x)\cos(x)}{\sin^4(x)}}{\normalsize 2x}\right)=\lim_{x\to0}\left(\frac{\sin(x)-x \cos(x)}{\sin^3(x)}\right)$$

Again we get the form $0/0$, so applying L'Hopital for the second time we get $$\displaystyle \lim_{x\to0}\left((\csc x)^2-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{\cos(x)-\cos(x)+x \sin(x)}{3\sin^2(x)\cos(x)}\right)= \\ =\frac{1}{3}\lim_{x\to0}\left(\frac{x}{\sin(x)}\right)\cdot\lim_{x\to0}\left({\frac{1}{\cos(x)}}\right)$$

Using the known limit $\displaystyle \lim_{x\to0}\frac{x}{\sin(x)}=1$ we eventually get $$\displaystyle \lim_{x\to0}\left((\csc x)^2-\frac{1}{x^2}\right)=\frac{1}{3}$$

Galc127
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$ (\csc x)^2-\frac{1}{x^2}=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x }$ ~ $\frac{x^2-\sin^2x}{x^4 }$

So using l'Hopital's rule

$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)=\lim\limits_{x\to0}\frac{2x-2\sin x\cos x}{4x^3}=\lim\limits_{x\to0}\frac{2x-\sin 2x}{4x^3}=\lim\limits_{x\to0}\frac{2-2\cos 2x}{12x^2}=\lim\limits_{x\to0}\frac{0+4\sin 2x}{24x}=\lim\limits_{x\to0}\frac{0+8\cos 2x}{24}=\frac13.$$

Paul
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  • Instead of twiddles, you can put the concrete factor $x^2/\sin^2x$, and separate it off before the l'Hopital – Empy2 Oct 23 '14 at 13:26