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I would like to calculate the following limit below.

How do you evaluate the limit?

$$\lim_{x\to\infty}{(1 + 2x)^\left(\frac{1}{2\ln x}\right)}$$

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    $$\large\lim_{\color{red}{n}\to\infty}{(1 + 2x)^\left(\frac{1}{2 \ln x}\right)}$$ – 2017-01-10

3 Answers 3

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With l'Hopital

Let $\displaystyle y=(1 + 2x)^{\frac{1}{2\ln x}}$ Then $\displaystyle \ln y=\frac{\ln(1 + 2x)}{2\ln x}$ and $$ \lim_{x\to\infty}\ln y=\lim_{x\to\infty}\frac{\ln(1 + 2x)}{2\ln x}=\lim_{x\to\infty}\frac{2x}{2(2x+1)}=\frac12 $$ then $y\to\sqrt{e}$.

Without l'Hopital

We know $\displaystyle\lim_{x\to\infty}\ln x=\infty$ and $\displaystyle 1+2x\sim 2x $ as $x\to\infty$, so $$ \lim_{x\to\infty}(1 + 2x)^{\frac{1}{2\ln x}}=\lim_{x\to\infty}(2x)^{\frac{1}{2\ln x}}=\lim_{x\to\infty}y$$ $$\ln y=\frac{\ln(2x)}{2\ln x}=\frac{\ln2+\ln x}{2\ln x}=\frac12+\frac{\ln 2}{2}\lim_{x\to\infty}\frac{1}{\ln x}=\frac12 $$ and $y\to\sqrt{e}$ again.

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    I'm not sure you can just switch $\lim (1+2x) = \lim (2x)$ when these limits are not finite. Surely you couldn't switch, for example, $\lim(1+2x)$ and $\lim x^n$. More justification is needed, is what I'm trying to say. For example, $\lim\frac{\ln(1+2x)}{\ln(2x)} = 1$ would be useful. – 2017-01-10
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    @Ennar This is equivalence like $\sin x\sim x$ as $x\to0$ and we write $1+2x\sim2x$ as $x\to\infty$ because we can ignore 1 beside $\infty$, in the other words $1+\infty=\infty$ informal. – 2017-01-10
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    I know what it is, the question is whether OP does. It does need justification, and $\lim (1+2x) = \lim(2x)\implies \lim(1+2x)^{1/(2\ln x)} = \lim(2x)^{1/(2\ln x)}$ is not correct argument. – 2017-01-10
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Let $L = \lim_{x\to\infty} (1+2x)^{\frac 1{2\ln x}}$. Then,

$$\ln L = \ln\lim_{x\to\infty} (1+2x)^{\frac 1{2\ln x}} = \lim_{x\to\infty} \frac {\ln(1+2x)}{2\ln x}$$ and you can use L'Hospital.

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We have $$\lim_{x \to \infty} (1+2x)^{\frac {1}{2\ln x}} = \lim_{x \to \infty} e^{\frac {\ln (1+2x)}{2 \ln x}} = e^{\frac {1}{2} \lim_{x \to \infty} \frac {\ln (1+2x)}{\ln x}} $$

Can you now continue using L'Hopital's rule? The answer is $\boxed {\sqrt {e}} $. Hope it helps.