# Evaluating $\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$ [duplicate]

Evaluate: $$\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$$

I have been trying to solve this for $15$ minutes but sin(sin(x)) part has me stuck.

My attempt:

I tried multiplying with $x$ inside the $\sin$ as $\sin{(\frac{x\sin{x}}{x})}$. No leads.

Use $\sin(u)=u-\frac{u^3}{6}+\frac{u^5}{120}+o(u^6)$ (three times).

• Netchaiev what is $o$? – prog_SAHIL Feb 2 '18 at 17:38
• it is a notation : if you have some quantity depending on $u$ (let us denote it by $f(u)$) such that $f(u)/u^4\rightarrow 0$ when $u\rightarrow 0$ (i.e. $f(u)$ is neglectful to respect as u^4 near $0$), you can denote $f(u)$ by $o(u^4)$ – Netchaiev Feb 2 '18 at 17:43
• Isn't there a simpler solution, because I am a high school-er. – prog_SAHIL Feb 2 '18 at 17:45
• I'm not sure about that ... The $\sin(\sin(x))$ is not the biggest issue : you can use $x\sin(\sin(x))=x\sin(x)\underbrace{\frac{\sin(\sin(x))}{\sin(x)}}_{\rightarrow 1}$, so $x\sin(\sin(x)) \sim x\sin(x)$, but it won't solve the indetermination (of order $x^6$) .... – Netchaiev Feb 2 '18 at 17:52
• Netchaiev thanks. – prog_SAHIL Feb 2 '18 at 18:00

Note that by Taylor's expansion

• $x\sin(\sin x)=x^2-\frac13x^4+\frac1{10}x^6+o(x^6)$
• $\sin^2x=x^2-\frac13x^4+\frac2{45}x^6+o(x^6)$

thus

$$\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}=\frac{x^2-\frac13x^4+\frac1{10}x^6-x^2+\frac13x^4-\frac2{45}x^6+o(x^6)}{x^6}=\frac1{18}+o(1)\to\frac1{18}$$

$$\sin(u)=u-\frac{u^3}{6}+\frac{u^5}{125}+o(u^6)$$

Then, $$\sin^2(x)=(x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6))^2 =x^2-\frac{x^4}{3} +o(x^6)$$ and

$$\sin(\sin(x))=\sin\left(x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6)\right) \\=x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6) -\frac{1}{6}\left(x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6)\right)^3 \\=x-\frac{x^3}{3}+\frac{x^5}{10}+o(x^5)+o(x^6)$$

Hence

$$\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}\\=\frac{x^2-\frac13x^4+\frac1{10}x^6-x^2+\frac13x^4-\frac2{45}x^6+o(x^6)}{x^6}=\frac1{18}+o(1)$$ Then the limit is $\frac1{18}$