Question Number 185405 by greougoury555 last updated on 21/Jan/23
$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({x}+\mathrm{2}\right)^{\mathrm{1}/{x}} −{x}^{\mathrm{1}/{x}} }{\left({x}+\mathrm{3}\right)^{\mathrm{1}/{x}} −{x}^{\mathrm{1}/{x}} }\:=? \\ $$
Answered by Frix last updated on 22/Jan/23
![=lim_(t→0^+ ) (((2t+1)^t −1)/((3t+1)^t −1)) = =lim_(t→0^+ ) (((d^2 [(2t+1)^t −1])/dt^2 )/((d^2 [(3t+1)^t −1])/dt^2 )) =(4/6)=(2/3)](Q185469.png)
$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\left(\mathrm{2}{t}+\mathrm{1}\right)^{{t}} −\mathrm{1}}{\left(\mathrm{3}{t}+\mathrm{1}\right)^{{t}} −\mathrm{1}}\:= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\frac{{d}^{\mathrm{2}} \left[\left(\mathrm{2}{t}+\mathrm{1}\right)^{{t}} −\mathrm{1}\right]}{{dt}^{\mathrm{2}} }}{\frac{{d}^{\mathrm{2}} \left[\left(\mathrm{3}{t}+\mathrm{1}\right)^{{t}} −\mathrm{1}\right]}{{dt}^{\mathrm{2}} }}\:=\frac{\mathrm{4}}{\mathrm{6}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$