How to evaluate $\lim _{x\to \infty }\left(1+2x\sqrt{x}\right)^{2/\ln x}$? I have a problem with this limit, can you please show me a way to solve it without L'Hôpital's rule?
$$\lim _{x\to \infty}\left(1+2x\sqrt{x}\right)^{\frac{2}{\ln x}}$$
This is my solution (it's correct, but I'd like to see another way, which doesn't use L'Hôpital's rule):
$$\lim _{x\to \infty}\left(e^{\frac{2}{\ln x}\ln\left(1+2x\sqrt{x}\right)}\right)$$
$$\frac{2\ln\left(1+2x\sqrt{x}\right)}{\ln x}$$
L'Hôpital:
$$\lim _{x\to \infty}\left(\frac{2\ln\left(1+2x\sqrt{x}\right)}{\ln x}\right)=\lim _{x\to \infty}\left(\frac{\frac{6\sqrt{x}}{2x^{\frac{3}{2}}+1}}{\frac{1}{x}}\right)=\lim _{x\to \infty}\left(\frac{6x^{\frac{3}{2}}}{2x^{\frac{3}{2}}+1}\right)$$
Again if you want:
$$=\lim_{x\to \infty}\left(\frac{9\sqrt{x}}{3\sqrt{x}}\right)=3=\color{red}{e^3}$$
Are there other ways to solve it?
 A: I think the key to make short work with this one is to recognize that the $+1$ in the brackets is neglible in the large $x$ limit.
Therefore we want to calculate
$$
\lim_{x\rightarrow\infty}(1+2x^{1/2}x)e^{\frac{2}{\log(x)}}=\lim_{x\rightarrow\infty}e^{2\frac{\log(2)}{\log(x)}+2\frac{\log(\sqrt{x})}{\log(x)}+2\frac{\log(x)}{\log(x)}}
$$
because $1/\log(x)\rightarrow 0$ as $x \rightarrow \infty$ it is also clear that the first term in exponent will drop out. Furthermore we may remember that $\log(x^{1/2})=\frac{1}{2}\log(x)$ and therefore
$$
\lim_{x\rightarrow\infty}e^{2\frac{\log(2)}{\log(x)}+2\frac{\log(\sqrt{x})}{\log(x)}+2\frac{\log(x)}{\log(x)}}={e^{2\left(\frac{1}{2}+1\right)}}=e^3
$$
Edit: u may be even quicker by using $x\sqrt{x}=x^{3/2}$ :-P
A: HINT:
$$\lim_{x\to\infty}\space\left(1+2x\sqrt{x}\right)^{\frac{2}{\ln(x)}}=$$
$$\lim_{x\to\infty}\space\exp\left[\ln\left(\left(1+2x\sqrt{x}\right)^{\frac{2}{\ln(x)}}\right)\right]=$$
$$\lim_{x\to\infty}\space\exp\left[\frac{2}{\ln(x)}\ln\left(1+2x\sqrt{x}\right)\right]=$$
$$\lim_{x\to\infty}\space\exp\left[\frac{2\ln\left(1+2x\sqrt{x}\right)}{\ln(x)}\right]=$$
$$\space\exp\left[2\lim_{x\to\infty}\frac{\ln\left(1+2x\sqrt{x}\right)}{\ln(x)}\right]$$
A: I'd first do $t=\sqrt{x}$ and compute the limit of the logarithm, because l'Hôpital is very simple in this case:
$$
\lim_{t\to\infty}\frac{\ln(2+t^3)}{\ln t}=
\lim_{t\to\infty}\frac{\dfrac{3t^2}{2+t^3}}{\dfrac{1}{t}}=
\lim_{t\to\infty}\dfrac{3t^3}{2+t^3}=3
$$
A: Apply $\ln$ to get
$$\tag 1 \frac{2}{\ln x}\, \ln (1+2x^{3/2}).$$
For $x>1,$ we have
$$\tag 2 \ln (2x^{3/2}) = \ln 2 + (3/2) \ln x < \ln (1+2x^{3/2}) < \ln (3x^{3/2}) =\ln 3 + (3/2) \ln x.$$
Multiply $(2)$ by $2/\ln x$ and apply the squeeze theorem to see that the limit of $(1)$ is $2\cdot (3/2) = 3.$ Exponentiating back gives $e^3$ for the desired limit.
