calculus limit question: another difficult limit problem

Question

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

Answer 1

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

Answer 2

$\frac 1{1-x^2}=1+x^2+O(x^4)$ 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 $\frac{23}{24}$.

Answer 3

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 $x \le 0$

Answer 4

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