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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?

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  • $\begingroup$ Another way could be using expansions, but it would be way more lengthy. $\endgroup$
    – Max Payne
    Commented Jan 28, 2016 at 11:45
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    $\begingroup$ You mean with Taylor or Asymptotic expansion? $\endgroup$
    – Amarildo
    Commented Jan 28, 2016 at 11:50

4 Answers 4

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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

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HINT:

$$\lim_{x\to\infty}\space\left(1+2x\sqrt{x}\right)^{\frac{2}{\ln(x)}}=$$ $$\lim_{x\to\infty}\space\exp\left[\ln\left(\left(1+2x\sqrt{x}\right)^{\frac{2}{\ln(x)}}\right)\right]=$$ $$\lim_{x\to\infty}\space\exp\left[\frac{2}{\ln(x)}\ln\left(1+2x\sqrt{x}\right)\right]=$$ $$\lim_{x\to\infty}\space\exp\left[\frac{2\ln\left(1+2x\sqrt{x}\right)}{\ln(x)}\right]=$$ $$\space\exp\left[2\lim_{x\to\infty}\frac{\ln\left(1+2x\sqrt{x}\right)}{\ln(x)}\right]$$

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  • $\begingroup$ i arrived here there, but now how to continue without Hopital? $\endgroup$
    – Amarildo
    Commented Jan 28, 2016 at 16:51
  • $\begingroup$ Taylor series, or just Hopital $\endgroup$ Commented Jan 28, 2016 at 16:52
  • $\begingroup$ you can show it with Taylor? Thanks $\endgroup$
    – Amarildo
    Commented Jan 28, 2016 at 16:53
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I'd first do $t=\sqrt{x}$ and compute the limit of the logarithm, because l'Hôpital is very simple in this case: $$ \lim_{t\to\infty}\frac{\ln(2+t^3)}{\ln t}= \lim_{t\to\infty}\frac{\dfrac{3t^2}{2+t^3}}{\dfrac{1}{t}}= \lim_{t\to\infty}\dfrac{3t^3}{2+t^3}=3 $$

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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.

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