Csdrhrt

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.

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

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

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 $x le 0$

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

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

Let $t=dfrac{x} {2(1-x^2)}$ to simplify typing. Note that $tto 0$ and $t/xto 1/2$ as $xto 0$. The numerator can be rewritten as $$2sin^2t-2cdotfrac{x^2}{4}=2left(sin t-frac {x} {2}right)left(sin t+frac{x}{2}right)=2ABtext{ (say)} $$ Clearly we have $$frac{B} {x} =frac{1}{2}+dfrac{sin t}{t}cdotfrac{t}{x}to frac {1}{2}+1cdotfrac{1}{2}=1$$ as $xto 0$. And $$frac{A} {x^3}=frac{sin t-t}{t^3}cdotfrac{t^3}{x^3}+frac{1}{2}cdotfrac{1}{1-x^2}to-frac{1}{6}cdotfrac{1}{8}+frac {1}{2}=frac{23}{48}$$ Therefore the given expression tends to $23/24$ as $xto 0$. In the above process we have used the standard limit $$lim_{tto 0}frac{sin t} {t} =1$$ and the limit $$lim_{tto 0}frac {sin t-t} {t^3}=-frac{1}{6}$$ which is easily proved via a single application of L'Hospital's Rule or via Taylor series for $sin t$.