# calculus limit question: another difficult limit problem

I have posted previously on a problem in a similar vein here: Limit evaluation: very tough question, cannot use L'hopitals rule

I believe this problem is very similar, but it has stumped me.

$$\lim_{x \to 0}\frac{1-\frac12 x^2 - \cos\left(\frac{x}{1-x^2}\right)}{x^4}$$

Really appreciate it if someone has some insight on this.This comes out to be indeterminate if one plugs in zero. Following the idea from the link above, I tried to recognize this as derivative evaluated at zero of a function, BUT I could not find the function, because I tried to make this all over x, so that means the function I would create would generate a rational type with x^3 on the bottom. I guess I should also try to look at some trig limit identities as well.

Hope someone out there can see how to navigate this problem.

P

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You should use Taylor formula to solve these kind of limits. – Emanuele Paolini Jan 3 '14 at 5:44
Is there a typo in your problem? Perhaps $$\lim_{x \to 0}\frac{1-\frac12x^2 - \cos(\frac{x}{1-x^2})}{x^4} ?$$ – Stephen Montgomery-Smith Jan 3 '14 at 5:54
Yes, you are correct, I missed the 1/2 in front of x^2 – Palu Jan 3 '14 at 5:56
Hi,I am using the taylor cosine series expansion. – Palu Jan 3 '14 at 5:57
@user99279: You should edit your question rather than leaving the correction in the comments. – user21820 Jan 3 '14 at 5:58

Like what Emanuele said, asymptotic expansions are useful for this kind of limits, and in fact better than L'Hopital (which fails miserably for some limits):

$\frac{x}{1-x^2} \in x + x^3 + O(x^5) \to 0$ as $x \to 0$

[We keep the error term so that at the end we know the error of the final approximation.]

$\cos( \frac{x}{1-x^2} ) \in 1 - \frac{1}{2} ( \frac{x}{1-x^2} )^2 + \frac{1}{24} ( \frac{x}{1-x^2} )^4 + O( ( \frac{x}{1-x^2} )^6 ) \\ \subset 1 - \frac{1}{2} (x+x^3+O(x^5))^2 + \frac{1}{24} (x+x^3+O(x^5))^4 + O(x^6) \text{ as } x \to 0 \\ \subset 1 - \frac{1}{2} x^2 - \frac{23}{24} x^4 + O(x^6) \text{ as } x \to 0$

[We can make the substitution into the Taylor expansion only because the input to $\cos$ tends to 0.]

$\frac{ 1 - \frac{1}{2} x^2 - \cos( \frac{x}{1-x^2} ) }{ x^4 } \in \frac{ \frac{23}{24} x^4 + O(x^6) }{ x^4 } = \frac{23}{24} + O(x^2) \to \frac{23}{24}$ as $x \to 0$

But you must make sure you understand the meaning of the Big-O notation and when and why they can be used. To make it more concrete, you can in many cases find explicit constants for bounds instead of using Big-O notation. For example:

$1 - \frac{1}{2} x^2 + \frac{1}{24} x^4 - \frac{1}{720} x^6 \le cos(x) \le 1 - \frac{1}{2} x^2 + \frac{1}{24} x^4$ [obtained by repeated differentiation and Mean-value theorem]

$x + x^3 \le \frac{x}{1-x^2} \le x + x^3 + 2 x^5$ for sufficiently small $x \ge 0$

$x + x^3 \ge \frac{x}{1-x^2} \ge x + x^3 + 2 x^5$ for sufficiently small $x \le 0$

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$\frac 1{1-x^2}=1+x^2+O(x^4)$ then $\frac x{1-x^2}=x+x^3+O(x^5)$ then $\cos\left(\frac x{1-x^2}\right)=1-\frac{x^2}{2}-x^4+\frac{x^4}{24}+O(x^5)=1-\frac{x^2}{2}-\frac{23x^4}{24}+O(x^5)$ the limit is $\frac{23}{24}$.

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Just expand $\cos\left(\frac{x}{1-x^2}\right)$ using series expansion and simplify a bit

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Ignore the terms which have the numerator greater than x^4 because that will become zero anyway – Aman May 30 '15 at 7:21