# Taking the limit of $\frac{1}{x^2} - \frac{1}{x\sin(x)}$ as $x$ approaches $0$

Plugging zero into $x$ gives me infinity-infinity which is indeterminate. I then try to multiply the function by $$\frac{\frac{1}{x^2} + \frac{1}{x\sin(x)}}{\frac1{x^2} + \frac1{x\sin(x)}}$$ which gives me another undeterminate and harder function... I know I need to use L'Hospital's rule, but I can't seem to find the right algebraic form to use that rule.

• Myles, I would like to state a general rule that Peter used. If you end up with two fractions of the form 1/0 - 1/0 then that's a indeterminate form. Usually, the best thing is two combine them through a common denominator!. Then retaking the limit results (most of the time) in a 0/0 situation to which you can apply L'Hospital. (Please don not confuse that with the Quotient Rule) – imranfat Aug 25 '13 at 20:00

## 2 Answers

Hint \begin{align}\frac{1}{{{x^2}}} - \frac{1}{{x\sin x}} &=\frac{x\sin x-x^2}{x^3\sin x}\\&= \frac{x}{{\sin x}}\frac{{\sin x - x}}{{{x^3}}}\end{align}

And $$\sin x=x-\frac{x^3}6+o(x^3)\tag 1$$

As it has been noted $(1)$ is a consequence of L'Hôpital's rule, if you may.

• I know how you found the taylor series for sin(x) but how did you set the first part equal to x/sinx * (sinx-x)/x^3? – Myles Aug 25 '13 at 20:01
• @Myles Common denominator, then factor up and down. – Pedro Tamaroff Aug 25 '13 at 20:03
• @PeterTamaroff: it would seem pretty simple to use L'Hospital with the way things are split up:\begin{align} \lim_{x\to0}\frac1{x^2}-\frac1{x\sin(x)} &=\lim_{x\to0}\frac{\sin(x)-x}{x^2\sin(x)}\\ &=\lim_{x\to0}\frac{x}{\sin(x)}\ \lim_{x\to0}\frac{\sin(x)-x}{x^3}\\ &=\lim_{x\to0}\frac1{\cos(x)}\ \lim_{x\to0}\frac{-\cos(x)}{6}\\ &=1\cdot\left(-\frac16\right) \end{align} – robjohn Aug 25 '13 at 20:11
• @robjohn Yes, of course. In fact what I write is an application of L'Hôpital. – Pedro Tamaroff Aug 25 '13 at 20:11

Peter uses Taylor series for the sine function. In case you are not familiar with those, you must use L'Hospital three times. (Usually, L'Hospital's Rule is dealt with at an earlier stage than series).

The numerator will give you $-\cos x$ and the denominator gives $6$. Now "plug in" $x→0$ to find the answer: $-\dfrac 16$