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

Question

$$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$

Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.

\begin{align} \lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex] &=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex] &=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex] &=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex] &=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex] &=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex] &=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex] &=-\frac{1}{4} \end{align}

Answer 1

Hint: Write the function as $$\frac{\sin^2(x)-x^2}{x^4}\times \frac{x^2}{\sin^2(x)}$$ Otherwise use the Talor's expantion if you know it.

Answer 2

$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)=\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\lim_{x \to 0}\frac{(\sin x-x)(\sin x+x)}{x^4}$$ $$=\lim_{x \to 0}\frac{(\sin x+x)}{x}\lim_{x \to 0}\frac{x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2x(\sin x-x)}{x^4}=\lim_{x \to 0}\frac{2(\sin x-x)}{x^3}$$ $$=\lim_{x \to 0}\frac{2(\cos x-1)}{3x^2}=\lim_{x \to 0}\frac{-2\sin x}{6x}=\frac{-1}{3}.$$

Answer 3

By l'Hopital we have

$$\lim_{x \to 0}\frac{1}{x^2} - \frac{1} {\sin^2 x} =\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^2\sin^2 x}$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{\sin 2x-2x}{2x\sin^2 x+x^2\sin 2x }$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{2\cos 2x-2}{2\sin^2 x+2x\sin 2x+2x\sin 2x +2x^2\cos 2x}$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-4\sin 2x}{2\sin 2 x+8x\cos 2x+4 \sin 2x+4x\cos 2x-4x^2\sin 2x}$$

$$\stackrel{H.R.}=\lim_{x \to 0}\frac{-8\cos 2x}{12\cos 2 x+8\cos 2x-16x \sin 2x-8x\sin 2x+4\cos 2x-8x\sin 2x-8x^2\cos2x}$$

$$=\lim_{x \to 0}\frac{-8\cos 2x}{24\cos 2 x-32x \sin 2x-8x^2\cos2x} =\frac{-8}{24-0-0}=-\frac13$$

Answer 4

As an alternative by Taylor expansion as $x\to 0$

$$\sin x = x -\frac16x^3 + o(x^3)\implies \sin^2 x = \left(x -\frac16x^3 + o(x^3)\right)^2=x^2-\frac13x^4+o(x^4)$$

we have

$$\frac{1}{x^2} - \frac{1} {\sin^2 x} =\frac{\sin^2 x-x^2}{x^2\sin^2 x}=\frac{x^2-\frac13x^4+o(x^4)-x^2}{x^2\left(x^2-\frac13x^4+o(x^4)\right)}=$$$$=\frac{-\frac13x^4+o(x^4)}{x^4+o(x^4)}=\frac{-\frac13+o(1)}{1+o(1)}\to -\frac13$$

Answer 5

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $\sin(x)$. Using the Taylor expansion:

$$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} +\mathcal{O}(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $\mathcal{O}(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $\dfrac{1}{\sin^2(x)}$:

$$\frac{1}{\sin^2(x)} = \frac{1}{x^2}\left[1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^{-2}$$

To expand the square brackets, we can use:

$$\frac{1}{(1+u)^2} = 1-2 u + 3 u^2 + \mathcal{O}(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)$. We have:

$$u^2 = \left[- \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^2 = \frac{x^4}{36} +\mathcal{O}(x^6)$$

Therefore:

$$\frac{1}{1+u}= 1-2 u + 3 u^2 +\mathcal{O}(u^3)= 1 + \frac{x^2}{3} + \frac{x^4}{15} +\mathcal{O}(x^6)$$

And we see that:

$$\frac{1}{\sin^2(x)} = \frac{1}{x^2} + \frac{1}{3} + \frac{x^2}{15} +\mathcal{O}(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $\dfrac{1}{\sin^4(x)}$ by squaring both sides of this expansion, like:

$$\lim_{x\to 0}\left[\frac{1}{\sin^4(x)}-\frac{1}{x^4} - \frac{2}{3 x^2}\right]= \frac{11}{45}$$

Answer 6

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$\sin x = x-\frac16 x^3+o(x^3) \implies \frac1{\sin x}=\frac 1x\left(1-\frac16x^2+o(x^2)\right)^{-1}=\frac1x+\frac16x+o(x)$$

therefore

$$\left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) =\left( {\frac{1}{x}} + {\frac{1} {\sin x} }\right) \left( {\frac{1}{x}} - {\frac{1} {\sin x} }\right)=$$ $$=\left(\frac2x+\frac16x+o(x)\right) \left( -\frac16x+o(x)\right) =-\frac13+o(1) \to -\frac13$$

Answer 7

$$\lim_{x\to0}\frac{(\sin{x}+x)(\sin{x}-x)}{x\sin{x}\cdot x\sin{x}}$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$\lim_{x\to0}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}$$

$$\lim_{x\to0}\frac{x-\tan{x}}{x^3}=-\frac{1}{3}$$

$$\lim_{x\to 0}\frac{e^x-1-x}{x^2}=\frac{1}{2}$$

So using this,

$$\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot \frac{(\sin{x}+x)}{x}\cdot \frac{(\sin{x}-x)}{x^3}$$

$$1\cdot2\cdot -\frac{1}{6}$$

$$-\frac{1}{3}$$

Answer 8

As noticed in the comments, we are allowed to proceed as follows

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

but we are not allowed to proceed as follows

$$\ldots=\lim_{x \to 0} \left( \frac{\sin x+x}{x\sin x} \right)\lim_{x \to 0}\left( \frac{\sin x-x}{x\sin x} \right)=\ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$\ldots=\lim_{x \to 0} \frac {\cos x+1} {\sin x+x\cos x}\cdot \lim_{x \to 0} \frac {\cos x-1} {\sin x+x\cos x}=\ldots$$

and the LHS limit, in the form $\frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

$$ \ldots=\lim_{x \to 0} (\cos x+1)\,\lim_{x \to 0} \frac {\cos x-1} {(\sin x+x\cos x)^2}=\ldots$$

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

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In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) = \lim_{x \to 0}\left(\frac{\sin^2 x-x^2}{x^4}\cdot\frac{x^2}{\sin^2 x}\right) \stackrel{?} = \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\ldots$$

and since, using l'Hopital for each part, we have

$$\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}=\lim_{x \to 0}\frac{\sin 2x-2x}{4x^3}=\lim_{x \to 0}\frac{2\cos 2x-2}{12x^2}=\lim_{x \to 0}\frac{-4\sin 2x}{24x}=\lim_{x \to 0}\frac{-8\cos 2x}{24}=-\frac13$$

$$\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\lim_{x \to 0}\frac{2x}{\sin 2x }=\lim_{x \to 0}\frac{2}{2\cos 2x }=1$$

we see that the initial step is allowed and then we can conclude that

$$\ldots= \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=-\frac13\cdot 1 =-\frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $\lim_{x \to 0}\frac{\sin x }x=1$.

Answer 9

Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

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What you can do instead (notice the asymmetry):

$$\lim_{x\to0}\frac{\sin^2x-x^2}{x^2\sin^2x}=\lim_{x\to0}\frac{\sin^2x-x^2}{x^4}=\lim_{x\to0}\frac{\sin x+x}{x}\lim_{x\to0}\frac{\sin x-x}{x^3} \\=2\lim_{x\to0}\frac{\cos x-1}{3x^2}=-2\lim_{x\to0}\frac{\sin x}{6x}=-\frac13.$$