Limit:Trig function

1. May 28, 2007

f(x)

1. The problem statement, all variables and given/known data

evaluate :
$$\lim_{x\longrightarrow 0 } \frac{cos(sin\ x) \ - \ cos \ x }{x^4}$$

What I've tried

I didnt go for L'Hopitals rule seeing the power of the denominator .
I tried working this one by using the expansion series of sin and cos -:

$$lim= 1-\frac{sin^2x}{2!} + \frac{sin^4x}{4!} \cdots - 1 + \frac{x^2}{2!}- \frac{x^4}{4!} \cdots$$

Dividing by x^4

$$= \frac{sin^2x}{x^2}. \frac{-1}{2x^2}+\frac{sin^4x}{24x^4}+\frac{x^2}{2x^4}-\frac{1}{24}$$

$$= \frac{1}{2x^2}$$

which gives ∞ as the answer..
Plz suggest a way to work this out
thx

Last edited: May 28, 2007
2. May 28, 2007

benorin

Apply l'Hospital's rule to the given limit sufficiently many times and you will get it.

3. May 28, 2007

VietDao29

The last line is wrong. It boils down to:

$$- \frac{\sin ^ 2 x}{x ^ 4} + \frac{1}{x ^ 2}$$, which is: $$\infty - \infty$$, not just ∞ as you pointed out, one of your signs is wrong.

-------------------------

L'Hopitals' rule is one way, and yours is another:

We have for x near 0:

Since, the denominator is x4, all power of x higher than 4 will be dropped.

$$\cos (\sin (x)) \approx 1 - \frac{\sin ^ 2 x}{2} + \frac{\sin ^ 4 x}{4!} \approx 1 - \frac{1}{2} \left( x - \frac{x ^ 3}{6} \right) ^ 2 + \frac{x ^ 4}{24} \approx 1 - \frac{1}{2} \left( x ^ 2 - \frac{1}{3} x ^ 4 \right) + \frac{x ^ 4}{24}$$

$$\cos (x) \approx 1 - \frac{x ^ 2}{2} + \frac{x ^ 4}{24}$$

Now, plug all to your expression, and the answer shouldn't be far off.
Can you go from here? :)

Last edited: May 28, 2007
4. May 29, 2007

f(x)

Thx for the help, Vietdao, but could you plz explain how you did the followin transformation-:

Isnt the above equal to 0 ? (using (sinx/x)=1), whereas the limit computed earlier by the expansion you have listed is 1/6. why dont they match ?

Thx again

5. May 29, 2007

VietDao29

Yes, of course. Since the denominator is x4, so any powers of x that are higher than 4 in the numerator when divided by the denominator (i.e, x4) will tend to 0 as x tend to 0. Right? So they do not affect the final result. So what we should do is to keep all the x's whose power are less than or equal to 4.

For x near 0, we have:

$$\sin x = x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!} - ...$$

$$\cos x = 1 - \frac{x ^ 2}{2} + \frac{x ^ 4}{4!} - ...$$

When x tends to 0, sin(x) also tends to 0, so, we have:

$$\cos (\sin x) = 1 - \frac{\sin ^ 2 x}{2} + \frac{\sin ^ 4 x}{4!} - \frac{\sin ^ 6 x}{6!} ...$$

Since the denominator is in x, we have to make one more change to sin(x), i.e convert them into x.

$$\cos (\sin x) = 1 - \frac{\left( x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!} - ... \right) ^ 2}{2} + \frac{\left( x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!} - ... \right) ^ 4}{4!} - \frac{\left( x - \frac{x ^ 3}{3!} + \frac{x ^ 5}{5!} - ... \right) ^ 6}{6!} + ...$$

Now, the trick is to drop all the powers of x that is higher than 4, the ones that are in red.

$$... = 1 - \frac{ x ^ 2 - \textcolor{red}{\frac{x ^ 6}{(3!) ^ 2} + \frac{x ^ {10}}{(5!) ^ 2} - ...} - \frac{x ^ 4}{3} + \textcolor{red}{ \frac{x ^ 6}{60} + ... } }{2} + \frac{x ^ 4 + \textcolor{red}{ ...} }{4!} - \textcolor{red}{\frac{x ^ 6 + ...}{6!}}$$

After dropping all the unnecessary terms, we have:

$$... \approx 1 - \frac{1}{2} \left( x ^ 2 - \frac{1}{3} x ^ 4 \right) + \frac{x ^ 4}{24}$$

Do the same to cos(x), we have:

$$\cos x = 1 - \frac{x ^ 2}{2} + \frac{x ^ 4}{4!} - \textcolor{red}{ \frac{x ^ 6}{6!} + ...} \approx 1 - \frac{x ^ 2}{2} + \frac{x ^ 4}{4!}$$

Nope, when x tends to 0, sin(x) / x will tend to 1, that does not mean that it's 1.

Factor out 1 / x2, we have:
$$\frac{1}{x ^ 2} \left( \frac{\sin ^ 2 x}{x ^ 2} - 1 \right)$$

Now, as x -> 0, the 1 / x2 will tend to infinity, whereas, sin2(x)/x2 will tend to 1. Note that, it only tends to 1, it's not equal to 1. So actually, it's the form $$0 \times \infty$$

Is everything clear now? :)

Last edited: May 29, 2007
6. May 29, 2007

f(x)

Yeah, that was perfect explanation, even better than what my professor's teach on board :)

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