What's the mistake I'm making in evaluating the below limit? $\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$

Question

Below is my solution for this limit. I get $0$, but the answer, is in fact, $\frac{1}{18}$. What's the mistake I'm making?

{From here: A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$}

$$\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$$ $$=\lim_{x \to 0}\frac{x\sin(\sin x)}{x^{6}}-\lim_{x \to 0}\frac{\sin^{2}x}{x^{6}}$$ $$=\lim_{x \to 0}\frac{x\sin(\sin x)}{x^{6}}\cdot\frac{\sin x}{\sin x}\cdot\frac{x}{x}-\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{x^4}$$ $$=\lim_{x \to 0}\frac{\sin(\sin x)}{\sin x}\cdot\frac{\sin x}{ x}\cdot\frac{x^2}{x^6}-\lim_{x \to 0}\frac{1}{x^4}$$ $$=\lim_{x \to 0}\frac{1}{x^4}-\lim_{x \to 0}\frac{1}{x^4} =\lim_{x \to 0}\left(\frac{1}{x^4}-\frac{1}{x^4}\right)=0$$

Be careful doing splitting and combining. $$\lim_{x \to a}[f(x) + g(x)] = \lim_{x \to a}f(x) + \lim_{x \to a}g(x)$$ only works when ALL limits exist (i.e., both of $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ have to exist). Namely, $$\lim_{x \to 0}\dfrac{1}{x^4}$$ doesn't exist, so your last step isn't justifiable. — Clarinetist, Dec 20 '16 at 03:58
The first step is wrong already. You can't split a limit into a difference of two other limits unless you first establish that both exist. — dxiv, Dec 20 '16 at 03:58
you can write the terms up to $x^6$ in the Taylor expansion of the numerator. Worthwhile practice. — Will Jagy, Dec 20 '16 at 03:58
Possible duplicate of A limit problem $\lim\limits_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$ — user5954246, Dec 20 '16 at 04:39
Even I gave a link to that question; this question requested users to point out what's wrong with MY SOLUTION for the limit. So it isn't a duplicate. :) @DenisKa — Panglossian Oporopolist, Dec 20 '16 at 13:45
Classical mistake: you split the sum, compute the limits indipendently and regroup, but the two separate limits do not exist. — , Dec 20 '16 at 17:30

Will Jagy · Accepted Answer · 2016-12-20T17:18:00.893

$$ \sin t = t - \frac{t^3}{6} + \frac{t^5}{120} \mp ... $$ $$ \sin \sin x = \sin x -\frac{\sin^3 x}{6} + \frac{\sin^5 x}{120} \mp $$ $$ \sin \sin x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} \mp ... \right) - \frac{\left( x - \frac{x^3}{6} \pm \right)^3}{6} + \frac{\left( x - \mp \right)^5}{120} \pm $$ $$ \sin \sin x = \left( x - \frac{x^3}{6} + \frac{x^5}{120} \mp ... \right) - \frac{\left( x^3 - \frac{x^5}{2} \pm \right)}{6} + \frac{ x^5 - \mp }{120} \pm $$ $$ \sin \sin x = x - \frac{x^3}{3} + \frac{x^5}{10} \mp $$ $$ x \sin \sin x = x^2 - \frac{x^4}{3} + \frac{x^6}{10} \mp $$ $$ \sin^2 x = x^2 - \frac{x^4}{3} + \frac{2 x^6}{45} \mp $$ $$ x \sin \sin x - \sin^2 x = \frac{x^6}{10} - \frac{2 x^6}{45} \pm $$ $$ x \sin \sin x - \sin^2 x = \frac{9x^6}{90} - \frac{4 x^6}{90} \pm $$ $$ x \sin \sin x - \sin^2 x = \frac{5x^6}{90} \pm $$ $$ x \sin \sin x - \sin^2 x = \frac{x^6}{18} + O( x^7) $$ Meanwhile, the function depicted is even, all exponents in the final expression must be even, so we can say the slightly stronger $$ x \sin \sin x - \sin^2 x = \frac{x^6}{18} + O( x^8) $$

What's the mistake I'm making in evaluating the below limit? $\lim_{x \to 0}\frac{x\sin(\sin x) - \sin^{2}x}{x^{6}}$

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