Evaluating $\lim_{x\to+\infty}\frac{(\sqrt{x+x^3}-x)\ln(\frac{\sqrt{4x+1}}{2\sqrt x +3})}{x\arctan x}$

Question

$$\lim_{x\to+\infty}\frac{(\sqrt{x+x^3}-x)\ln(\frac{\sqrt{4x+1}}{2\sqrt x +3})}{x\arctan x}$$ I think,it would be useful to start with simplifying the first term in the numerator as $(\sqrt{x+x^3}-x)=x^{{3\over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $\ln1=0$

Answer 1

$$\lim_{x\to+\infty}\frac{(\sqrt{x+x^3}-x)\ln\frac{\sqrt{4x+1}}{2\sqrt x +3}}{x\arctan x}=\lim_{x\to+\infty}\frac{x^3\ln\left(\frac{\sqrt{4x+1}}{2\sqrt x +3}-1+1\right)}{(\sqrt{x+x^3}-x)x\arctan x}=$$ $$=\lim_{x\to+\infty}\frac{x^2\left(\frac{\sqrt{4x+1}}{2\sqrt x +3}-1\right)}{(\sqrt{x+x^3}-x)\arctan x}=\lim_{x\to+\infty}\frac{x^2(4x+1-4x-12\sqrt{x}-9)}{(2\sqrt x +3)(\sqrt{4x+1}+2\sqrt x +3)(\sqrt{x+x^3}-x)\arctan x}=$$ $$=\lim_{x\to+\infty}\frac{-12-\frac{8}{\sqrt{x}}}{\left(2 +\frac{3}{\sqrt{x}}\right)\left(\sqrt{4+\frac{1}{x}}+2 +\frac{3}{\sqrt{x}}\right)\left(\sqrt{\frac{1}{x^2}+1}-\frac{1}{\sqrt{x}}\right)\arctan x}=$$ $$=\frac{-12}{2\cdot4\cdot1\cdot\frac{\pi}{2}}=-\frac{3}{\pi}.$$

Answer 2

$$I=\lim_{x\to+\infty}\frac{(\sqrt{x+x^3}-x)\ln(\frac{\sqrt{4x+1}}{2\sqrt x +3})}{x\arctan x}=\dfrac2\pi\lim_{x\to+\infty}\frac{(\sqrt{x+x^3}-x)\ln(\frac{\sqrt{4x+1}}{2\sqrt x +3})}x$$

Set $x=1/h^2,h\to0^+$

$$\sqrt{x+x^3}-x=\sqrt{\dfrac1{h^2}+\dfrac1{h^6}}-\dfrac{1}{h^2}=\dfrac{\sqrt{1+h^4}-h}{h^3}\text{ and }\dfrac{\sqrt{4x+1}}{2\sqrt{x}+3}= \dfrac{\sqrt{4+h^2}}{2+3h} $$

$$\dfrac{\pi I}2=\lim_{h\to0^+}\dfrac{h^2(\sqrt{1+h^4}-h)}{h^3}\ln\left(\dfrac{\sqrt{4+h^2}}{2+3h}\right)$$

$2\ln\left(\dfrac{\sqrt{4+h^2}}{2+3 h}\right)=\ln\dfrac{4+h^2}{(2+3h)^2}=\ln\left(1-\dfrac{12h}{(2+3h)^2}\right)$

$$\pi I=-\lim_{h\to0^+}(\sqrt{1+h^4}-h)\cdot\lim_{h\to0^+}\left(\lim_{h\to0^+}\dfrac{\ln\left(1-\dfrac{12h}{(2+3h)^2}\right)}{-\dfrac{12h}{(2+3h)^2}}\right)\cdot\lim_{h\to0^+}\dfrac{\dfrac{12h}{(2+3h)^2}}h=-1\cdot1\cdot\dfrac{12}4$$