Evaluate $\lim_{x \to 0} \frac{1-(x^2/2) -\cos (x/(1-x^2))}{x^4}$

Question

find the limits with Using :$\lim_{x \to 0} \frac{1-\cos x}{x^2}=\frac12$

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

My Try :

$$\lim_{x \to 0} \frac{1-\dfrac{x^2}2-\cos (\dfrac{x}{1-x^2})}{x^4}=\lim_{x \to 0} \frac{(1-\cos (\dfrac{x}{1-x^2}))+(-\dfrac{x^2}2)}{x^4}$$

$$\lim_{x \to 0} \frac{(\dfrac{(1-\cos (\dfrac{x}{1-x^2}))}{(\dfrac{x}{1-x^2})^2})(\dfrac{x}{1-x^2})^2+(-\dfrac{x^2}2)}{x^4}$$

now what ?

Answer 1

Use the famous formula $1-\cos 2t=2\sin^{2}t$ to transform the expression into $$2\cdot\frac{\sin^{2}t-(x/2)^{2}}{x^{4}}$$ where $t=x/(2(1-x^{2}))$. Next the numerator needs to be split by adding and subtracting $t^2$. It is easy to check that $(t^2-(x/2)^{2})/x^4\to 1/2$ and hence the desired limit is equal to $$1+2\lim_{x\to 0}\frac{\sin^{2}t-t^2}{t^{4}}\cdot(t/x)^{4}$$ Since $t/x\to 1/2$ we can see that he desired limit is equal to $$1+\frac{1}{8}\lim_{t\to 0}\frac{\sin t +t} {t} \cdot\frac{\sin t-t} {t^3}$$ And then the answer is easily seen to be $1+(1/8)(2)(-1/6)=23/24$. Note that the last limit (which evaluates to $-1/6$) necessitates the use of tools like Taylor series or L'Hospital's Rule.

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The limit can not be solved just using $\lim\limits_{x\to 0}\dfrac{1-\cos x} {x^{2}}=\dfrac{1}{2}$ and algebraic manipulation.

Answer 2

Letting $u = x/(1 - x)^2$, we can write this as:

$$\lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u) + \frac{1}{2}(u^{2} - x^{2})}{u^{4}(1 - x^{2})^{4}}$$ $$= \lim_{u \rightarrow 0} \frac{1 - \frac{1}{2}u^{2} - \cos(u)}{u^{4}} + \lim_{x \rightarrow 0}\frac{u^{2} - x^{2}}{2x^{4}}$$ The first limit is a bit more complicated than the limit it said to use- but it's easy with Taylor series or l'Hospital's rule, and the answer is $-\frac{1}{24}$. The second limit is just a rational function in $x$: $$\lim_{x \rightarrow 0}\frac{u^2 - x^2}{2x^{4}} = \lim_{x \rightarrow 0} \frac{x^{2} - x^{2}(1 - x^{2})^{2}}{2x^{4}(1 - x^{2})^{2}} = \lim_{x \rightarrow 0}\frac{x^{2} - x^{2} + 2x^{4} - x^{6}}{2x^{4}(1 - x^{2})^{2}} = \lim_{x \rightarrow 0}\frac{2 - x^{2}}{2(1 - x^{2})^{2}} = 1$$

Answer 3

$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(\frac{x}{1-x^2}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}$$

Now for Taylor's expansion we have that

$$\lim\limits_{x \to 0}\frac{1}{1-x^2}=1-x^2+o(x^2)$$

So

$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x{\frac{1}{1-x^2}}\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x(1-x^2+o(x^2)\right)}{x^4}=$$

$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x-x^3+o(x^3)\right)}{x^4}$$

Now recall that for Taylor's expansion

$$\lim\limits_{x \to 0}\cos t=1-\frac{t^2}{2}+\frac{t^4}{24}+o(t^4)$$

where, in our case, $t=(x-x^3)$.

Hence

$$\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-\cos \left(x-x^3+o(x^3)\right)}{x^4}=\lim\limits_{x \to 0}\frac{1-\frac{x^2}{2}-1+\frac{x^2}{2}+x^4-\frac{x^4}{24}+o(x^4)}{x^4}=$$

$$=\lim\limits_{x \to 0}\frac{\frac{23}{24}x^4+o(x^4)}{x^4}=\frac{23}{24}.$$