Calculate $\lim_{x \rightarrow 0^+} \frac{\arctan (\log (1+\sqrt x)) \sin^3(x^{3/4})}{(e^{\tan(x)}-1)(1-\sin^2(x))}$

Question

I'm trying to calculate the following limit:

$$\lim_{x \rightarrow 0^+} \frac{\displaystyle\arctan (\log (1+\sqrt x)) \sin^3(x^\frac34)}{\displaystyle (e^{\tan(x)}-1)(1-\sin^2(x))}$$

For WolframAlpha the result is: $0$.

I did those steps, using Mac-Laurin:

$$e^{\tan(x)}=1+x+\frac{x^2}{2}+ o(x^2)$$ $$\sin^2(x)=x^2+ o(x^2)$$ Hence, the denominator became: $$x+ \frac{x^2}{2}+ o(x^2)$$ Then, I'm having issues with numerator: $$\arctan (\log (1+\sqrt x)) = \sqrt x - \frac {x}{2} + o(x^2)$$ $$\sin^3(x^\frac34)=x^{\frac94}+ o(x^3)$$

Someone could say me how to deal with the Numerator, o give me a hint for solve it? Thank you.

Answer 1

Note that by standard limits

$$\frac{\arctan (\log (1+\sqrt x))\cdot \sin^3(x^\frac34)}{(e^{\tan x}-1)\cdot (1-\sin^2 x)}=\frac{\arctan ( \log (1+\sqrt x) )}{ \log (1+\sqrt x) }\cdot\frac{ \log (1+\sqrt x)}{\sqrt x}\cdot \frac{\sin^3(x^\frac34)}{x^\frac94}\cdot \frac{\tan x}{e^{\tan x}-1}\cdot\frac{x}{\tan x}\cdot \frac{\sqrt x\cdot x^\frac94 }{x(1-\sin^2 x) }\to1\cdot1\cdot1\cdot1\cdot1\cdot0=0$$

indeed

$$\frac{\sqrt x\cdot x^\frac94 }{x(1-\sin^2 x) }=\frac{x^\frac74 }{(1-\sin^2 x) }\to \frac01=0$$

Answer 2

Near 0,

$$\ln (1+X)\sim X $$

$$\arctan (Y)\sim Y $$

$$e^Z-1\sim Z \sim \tan (Z)$$

Answer 3

The term $1-\sin^2x$ in the denominator can be disregarded because its limit is $1$.

Thus the denominator is $$ e^{\tan x}-1=1+\tan x+o(\tan x)-1=\tan x+o(\tan x)=x+o(x) $$

For the numerator, you can set $t=\sqrt[4]{x}$, so you have $$ \arctan(\log(1+t^2))=\arctan(t^2+o(t^2))=t^2+o(t^2) $$ and $$ \sin^3(t^3)=(t^3+o(t^3))^3=t^9+o(t^9) $$ Thus your final expansion is $$ (t^2+o(t^2))(t^9+o(t^9))=t^{11}+o(t^{11})=x^{11/4}+o(x^{11/4}) $$ Therefore your limit is $$ \lim_{x\to0^+}\frac{x^{11/4}+o(x^{11/4})}{x+o(x)}=0 $$