3

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.

Guy Fsone
  • 23,903
prog_SAHIL
  • 2,543
  • 13
  • 27

3 Answers3

4

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

Then, $$\sin^2(x)=(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6))^2 =x^2-\frac{x^4}{3} +o(x^6)$$ and

$$\sin(\sin(x))=\sin\left(x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6)\right) \\=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6) -\frac{1}{6}\left(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)\right)^3 + \frac{1}{120} \left(x + o(x^3)\right)^5 \\=x-\frac{x^3}{3}+\frac{x^5}{10}+o(x^6) $$

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 $\frac1{18}$

muzzlator
  • 7,325
Guy Fsone
  • 23,903
3

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

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

user
  • 154,566
2

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

Netchaiev
  • 4,811
  • Netchaiev what is $o$? – prog_SAHIL Feb 02 '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 02 '18 at 17:43
  • Isn't there a simpler solution, because I am a high school-er. – prog_SAHIL Feb 02 '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 02 '18 at 17:52
  • Netchaiev thanks. – prog_SAHIL Feb 02 '18 at 18:00
  • Can you please link a few sources where I can study about this, – prog_SAHIL Feb 02 '18 at 18:20
  • maybe this one http://www.math.ubc.ca/~feldman/m120/taylorLimits.pdf}, or here a video : https://fr.coursera.org/learn/advanced-calculus/lecture/OmGzx/how-do-taylor-series-provide-intuition-for-limits – Netchaiev Feb 02 '18 at 18:33
  • You can also compute Taylor series thanks to Wolfram Alpha : https://www.wolframalpha.com/input/?i=taylor+sin(x) – Netchaiev Feb 02 '18 at 18:36