Find $\lim_{x \to 0} \frac{x^6(1-\cos(x^6))}{(x-\sin(x))^6}$

Question

Find $$\lim_{x \to 0} \frac{x^6(1-\cos(x^6))}{(x-\sin(x))^6}.$$

I have tried using L'Hospital but I don't know if that's the easier way (or even if it would actually work), because it keeps having an indetermination. The result is 23328. I've also tried using $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, but haven't found a way.

Answer 1

(As we'll see below,) both the numerator and denominator have zeros of order $18$, so while L'Hôpital's Rule will work, you'll need to apply it $18$ times to compute the answer.

Hint Instead, applying Taylor's Theorem to $$\sin x = x - \frac{1}{3!} x^3 + \cdots \qquad \textrm{and} \qquad \cos (x^6) = 1 - \frac{1}{2} (x^{6})^2 + \cdots$$ gives that numerator is $$x^6 \left[1 - \left(1 - \frac{1}{2} (x^6)^2 + \cdots\right)\right] = x^{18} \left(\frac{1}{2} + R(x) \right)$$ for some function $R$ such that $R(0) = 0$, and that the denominator is $$\left[x - \left(x - \frac{1}{3!} x^3 + \cdots\right)\right]^6 = x^{18} \left(\frac{1}{3!^6} + S(x)\right) ,$$ for some function $S$ such that $S(0) = 0$.

Thus, the limit is $$\lim_{x \to 0} \frac{x^6(1-\cos(x^6))}{(x-\sin(x))^6} = \lim_{x \to 0} \frac{x^{18} \left(\frac{1}{2} + R(x) \right)}{x^{18} \left(\frac{1}{3!^6} + S(x)\right)} = \lim_{x \to 0} \frac{\frac{1}{2} + R(x)}{\frac{1}{3!^6} + S(x)} = \frac{\frac{1}{2}}{\frac{1}{3!^6}} = \frac{3!^6}{2} = \boxed{23\,328}$$ as claimed.

Answer 2

Use taylors expansion, first two terms $\cos(x)=1-\frac{x^{2}}{2}$ and $\sin (x)=x-\frac{x^{3}}{6}$.

$$\lim_{x \to 0}\frac{x^{6}(1-\cos (x^{6}))}{(x-\sin (x))^{6}}= \lim_{x \to 0}\frac{x^{6}(1-(1-\frac{x^{12}}{2}))}{(x-(x-\frac{x^{3}}{6}))^{6}}=23328$$ Hope I did help.

Answer 3

L'Hospital approach. We have that $$\frac{x^6(1-\cos(x^6))}{(x-\sin(x))^6}=\frac{1-\cos(x^6)}{(x^6)^2}\cdot \left(\frac{x^3}{x-\sin(x)}\right)^6.$$ By L'Hospital's Rule, $$\lim_{t\to 0}\frac{1-\cos(t)}{t^2}=\lim_{t\to 0}\frac{\sin(t)}{2t}=\frac{1}{2}\implies \lim_{x\to 0}\frac{1-\cos(x^6)}{(x^6)^2}=\frac{1}{2}$$ and $$\lim_{x\to 0}\frac{x^3}{x-\sin(x)}=\lim_{x\to 0}\frac{3x^2}{1-\cos(x)}=\frac{3}{1/2}=6.$$ Therefore $$\lim_{x\to 0}\frac{x^6(1-\cos(x^6))}{(x-\sin(x))^6}= \lim_{x\to 0}\frac{1-\cos(x^6)}{(x^6)^2}\cdot \left(\lim_{x\to 0}\frac{x^3}{x-\sin(x)}\right)^6=\frac{6^6}{2}=23328.$$