Evaluating a limit $\lim_{x\to 0}\left\{\dfrac 2{x^3}(\tan x- \sin x )\right\}^{2/x^2 }$

Question

Find the value of: $$\lim_{x\to 0}\left\{\dfrac 2{x^3}(\tan x- \sin x )\right\}^{2/x^2 }$$

The bracketed part is in $0/0 $ indeterminate form and the exponent is tending to infinity. So I would like to know the general method of evaluating such limits of $\left(\dfrac{0}{0}\right)^\infty$ form.

Any hints on how to proceed with this question?

Answer 1

Hint:

Determine first the limit of $$\ln\left(\frac 2{x^3}(\tan x- \sin x )\right)^{\tfrac2{x^2}}=\frac 2{x^2}\ln\left(\frac 2{x^3}(\tan x- \sin x )\right)$$ For that, you need to expand $\tan x-\sin x$ at order $5$: $$\tan x-\sin x=x+\frac {x^3}3+\frac{2}{15}x^5+o(x^5)-\Bigl(x-\frac {x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)=\frac {x^3}2+\frac{x^5}{8}+o(x^5),$$ $$\text{so }\qquad\ln\left(\frac 2{x^3}(\tan x- \sin x )\right)=\ln\Bigl(1+\frac{x^2}4+o(x^2)\Bigr)=\frac{x^2}4+o(x^2). $$ $$\text{and finally }\quad\frac 2{x^2}\,\ln\left(\frac 2{x^3}(\tan x- \sin x )\right)= \frac12+o(1)\to\frac12.$$

As a conclusion, the limit of the given expression is $\;\sqrt e$.

Answer 2

Since $\sin x\approx x-\frac{x^3}{6}+\frac{x^5}{120}$ and $\tan x\approx x+\frac{x^3}{3}+\frac{2x^5}{15}$, $\frac{2}{x^3}(\tan x-\sin x)\approx \frac{2}{x^3}(\frac{x^3}{3}+\frac{2x^5}{15}+\frac{x^3}{6}-\frac{x^5}{120})=1+\frac{x^2}{4}\approx\exp\frac{x^2}{4}$. The limit is therefore $\sqrt{e}$.

Answer 3

Let$$ y=\lim_{x\to 0}\left\{\dfrac 2{x^3}(\tan x- \sin x )\right\}^{2/x^2}$$$$$$ $$\Rightarrow ln(y)=\lim_{x\to 0}\left(\dfrac{2}{x^2}\right)\ln\left(\dfrac 2{x^3}(\tan x- \sin x )\right)$$ $$\Rightarrow ln(y)=\lim_{x\to 0}\left(\dfrac{2}{x^2}\right)\ln\left(\dfrac 2{x^3}\left(\left(x+\frac{x^3}3+\frac{2x^5}{15}+...\right)-\left(x-\frac{x^3}6+\frac{x^5}{120}+...\right) \right)\right)$$

$$\Rightarrow ln(y)=\lim_{x\to 0}\left(\dfrac{2}{x^2}\right)\ln\left(\frac{2}{x^3}\left(\frac{x^3}2+\frac{x^5}8+\frac{13x^7}{240}+...\right)\right)$$

$$\Rightarrow ln(y)=\lim_{x\to 0}\left(\dfrac{2}{x^2}\right)\ln\left(1+\frac{x^2}4+\frac{13x^4}{120}+...\right)$$ $$$$ Now recall that $$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3+... $$ $$$$ Hence, $$\ln\left(1+\frac{x^2}4+\frac{13x^4}{120}+...\right)=\left(\frac{x^2}4+\frac{13x^4}{120}+...\right)-\dfrac{\left(\frac{x^2}4+\frac{13x^4}{120}+...\right)}{2}+...$$ $$$$ Hence, $$\lim_{x\to 0}\left(\dfrac{2}{x^2}\right)\ln\left(1+\frac{x^2}4+\frac{13x^4}{120}+...\right)=\lim_{x\to 0}\left(\dfrac{2}{x^2}\right)\left(\left(\frac{x^2}4+\frac{13x^4}{120}+...\right)-\dfrac{\left(\frac{x^2}4+\frac{13x^4}{120}+...\right)}{2}+...\right)=\frac12$$ $$$$ $$\Rightarrow ln(y)=\frac12\Rightarrow y=e^{\frac12}$$

Answer 4

By Taylor's expansion we have that

• $\tan x=x+\frac{x^3}3+\frac{2x^5}{15}+o(x^5)$

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

• $\tan x -\sin x=\frac{x^3}2+\frac{x^5}8+o(x^5)$

then

$$\left(\dfrac 2{x^3}(\tan x- \sin x )\right)^{2/x^2 }=\left(1+\frac{x^2}4+o(x^2)\right)^{2/x^2 }=\\=\left[\left(1+\frac{x^2}4+o(x^2)\right)^{\frac1{\frac{x^2}4+o(x^2)} }\right]^{\frac{\frac{x^2}4+o(x^2)}{\frac{x^2}2}}\to [e]^\frac12=\sqrt e$$