# Find the limit using Maclaurin series:

## Homework Statement

$$\lim_{x\to0}[\frac{1}{x^2} - \frac{\cos(x)}{\sin(x)^2}]$$

I'm supposed to use Maclaurin series to evaluate this limit. The instructions suggest, as a hint:

"First combine the fractions. Then find the first term of the denominator series and the first term of the numerator series".

## Homework Equations

- Maclaurin series for sin(x)
- Maclaurin series for cos(x)
- L'Hopital's rule

## The Attempt at a Solution

Graphing it I know the limit is (-1/6), but I can't show it.

I tried separating each function into its respective series, and I've tried taking several derivatives to see if L'Hopital's rule can make the limit apparent, but I'm not having much luck. I've also tried using different trig identities to try to help, but without much luck.

Any suggestions?

Mark44
Mentor

## Homework Statement

$$\lim_{x\to0}[\frac{1}{x^2} - \frac{\cos(x)}{\sin(x)^2}]$$

I'm supposed to use Maclaurin series to evaluate this limit. The instructions suggest, as a hint:

"First combine the fractions. Then find the first term of the denominator series and the first term of the numerator series".

## Homework Equations

- Maclaurin series for sin(x)
- Maclaurin series for cos(x)
- L'Hopital's rule

## The Attempt at a Solution

Graphing it I know the limit is (-1/6), but I can't show it.

I tried separating each function into its respective series, and I've tried taking several derivatives to see if L'Hopital's rule can make the limit apparent, but I'm not having much luck. I've also tried using different trig identities to try to help, but without much luck.

Any suggestions?
Did you try the first thing stated in the hint? "First combine the fractions."

Yeah, but that didn't seem to help. I'm left with a jumble of series and not really sure where to go from there.

It says to take the first term of both numerator and denominator, but given that the denominator is the square of the series (and that the first term is zero), I'm not really sure. And the numerator has several series...

Mute
Homework Helper
If you plug in the taylor series for cos and sin you get

$$\frac{1 - \frac{x^2}{2} + \mathcal{O}(x^4)}{(x - x^3/6 + \mathcal{O}(x^5))^2} = \frac{1}{x^2}\frac{1 - \frac{x^2}{2} + \mathcal{O}(x^4)}{(1 - x^2/6 + \mathcal{O}(x^4))^2}$$

Expand the denominator using the binomial expansion (treat the $(x^2/6 + \mathcal{O}(x^4))^2$ stuff as the small "x" in 1/(1-x) = 1 + x + ...):

$$\frac{1}{x^2}\frac{1 - \frac{x^2}{2} + \dots}{1 - x^2/6 + \mathcal{O}(x^4)} = \frac{1}{x^2}(1 - \frac{x^2}{2} + mathcal{O}(x^4))(1 + (x^2/6 + \mathcal{O}(x^4))^2 - (x^2/6 + \mathcal{O}(x^4)))^4 + \dots)$$

Now multiply things through, keeping the few lowest order terms. Your goal is to get something that looks like $1/x^2 - \mbox{constant} + \mathcal{O}(x)$. Since you already have a 1/x^2, it cancels the 1/x^2 term and you're left with your answer.

vela
Staff Emeritus
$$\sin^2 x = \left(x - \frac{x^3}{3!} + \cdots\right)\left(x - \frac{x^3}{3!} + \cdots\right)$$
$$\sin^2 x = x^2 + \left[-x\times\frac{x^3}{3!} -\frac{x^3}{3!}\times x\right] + \cdots = x^2 - \frac{x^4}{3} + \cdots$$