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Find the limit of :

$$\lim\limits_{x \to 0^+}{(2\sqrt{x}+x)^\frac{1}{\ln x}}{}$$

I've tried to make it look like an exponent of e:

$$e^\frac{\ln (2*\sqrt{x}+x)}{\ln x}$$

but, then again I reach an indeterminate form of infinity divides by infinity.

I then tried to use L'Hospital's rule, which also seems not to work.

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According to l'Hopital, $$\lim_{x\to0^+}\frac{\ln(2\sqrt x+x)}{\ln x}=\lim_{x\to 0^+}\frac{(x^{-1/2}+1)\frac1{2\sqrt x+x}}{\frac1x} $$ if the latter exists. But $$\lim_{x\to0^+}\frac{(x^{-1/2}+1)\frac1{2\sqrt x+x}}{\frac1x} =\lim_{x\to0^+}\frac{{\sqrt x+x}}{2\sqrt x+x}=\lim_{x\to0^+}\frac{1+\sqrt x}{2+\sqrt x}=\frac12. $$

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Enforce the substitution $x=e^t$. Then, we have

$$\begin{align} \lim_{x\to0^+}\left(2\sqrt x+x\right)^{1/\log(x)}&=\lim_{t\to-\infty}\left(2e^{t/2}+e^t\right)^{1/t}\\\\ &=e^{1/2}\lim_{t\to-\infty}\left(2+e^{t/2}\right)^{1/t}\\\\ &=e^{1/2} \end{align}$$

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  • $\begingroup$ Nice approach, but is "enforce" really the right word here? $\endgroup$ – hmakholm left over Monica Jun 20 '18 at 13:50
  • $\begingroup$ @HenningMakholm Thank you! To enforce means "to cause (something) to happen." So, it is one of many "right" words. $\endgroup$ – Mark Viola Jun 21 '18 at 2:46
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Note $x^{1/\ln x} = e,$ which follows by applying $\ln$ to both sides. Thus $\sqrt x^{1/\ln x} = e^{1/2}.$

Now $x<\sqrt x$ for $0<x<1,$ and because the power $1/\ln x<0,$ we have

$$(3\sqrt x)^{1/\ln x} < (2\sqrt x+x)^{1/\ln x} < (2\sqrt x)^{1/\ln x}$$

for $x$ in this range. The left side equals $3^{1/\ln x}e^{1/2}.$ As $x\to 0^+,$ the limit of this is $3^0\cdot e^{1/2}= e^{1/2}.$ Similarly the right side $\to e^{1/2}.$ Therefore $e^{1/2}$ is the desired limit.

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L'Hospital is not necessary.

$$(2\sqrt x+x)^{1/\ln x}=(2+\sqrt x)^{1/\ln x}\sqrt x^{1/\ln x}.$$

The first factor tends to $(2+0)^0=1$, and the second to $e^{\ln x/2\ln x}=\color{green}{\sqrt e}$.

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