Desirae Washington

2022-07-13

Evaluate $\underset{x\to \mathrm{\infty }}{lim}{\left(\frac{1}{x}+{2}^{\frac{1}{x}}\right)}^{x}$

Feriheashrz

Expert

You have that
${e}^{1+x{2}^{\frac{1}{x}}-x}={e}^{1+x\left({2}^{\frac{1}{x}}-1\right)}$
Now recall that
$\underset{t\to 0}{lim}\frac{{a}^{t}-1}{t}=\underset{t\to 0}{lim}\frac{{e}^{t\mathrm{log}a}-1}{t}=\underset{t\to 0}{lim}\frac{{e}^{t\mathrm{log}a}-1}{t\mathrm{log}a}\cdot \mathrm{log}a$
By putting $u=t\mathrm{log}a$ and noticing that $t\mathrm{log}a\to 0$ as $t\to 0$, hence
$\underset{t\to 0}{lim}\frac{{e}^{t\mathrm{log}a}-1}{t\mathrm{log}a}\cdot \mathrm{log}a=\underset{u\to 0}{lim}\frac{{e}^{u}-1}{u}\mathrm{log}a=1\cdot \mathrm{log}a=\mathrm{log}a$
So, in ${e}^{1+x\left({2}^{\frac{1}{x}}-1\right)}$, putting $z=\frac{1}{x}$ you have that $z\to 0$ as $x\to \mathrm{\infty }$ and so
$\underset{x\to \mathrm{\infty }}{lim}{e}^{1+x\left({2}^{\frac{1}{x}}-1\right)}=\underset{z\to {0}^{+}}{lim}{e}^{1+\frac{{2}^{z}-1}{z}}={e}^{1+\mathrm{log}2}=e\cdot {e}^{\mathrm{log}2}=2e$
Alternatively, notice that since when $x\to \mathrm{\infty }$ it is $\frac{1}{x}+{2}^{1/x}>0$, it is
${\left(\frac{1}{x}+{2}^{1/x}\right)}^{x}=\mathrm{exp}\left[x\mathrm{log}\left(\frac{1}{x}+{2}^{1/x}\right)\right]=\mathrm{exp}\left\{x\left[\frac{1}{x}\mathrm{log}2+\mathrm{log}\left(1+\frac{1}{x\cdot {2}^{1/x}}\right)\right]\right\}$
Now use that $\mathrm{log}\left(1+t\right)=t+\text{o}\left(t\right)$ as $t\to 0$; I'm using $\mathrm{exp}\left[f\left(x\right)\right]={e}^{f\left(x\right)}$ for better formatting.

Shea Stuart

Expert

Taking the logarithm of the function and setting $t:=\frac{1}{x}$, we can use L'Hospital:
$\underset{t\to 0}{lim}\frac{\mathrm{log}\left(t+{2}^{t}\right)}{t}=\underset{t\to 0}{lim}\frac{\left(1+{2}^{t}\mathrm{log}2\right)}{t+{2}^{t}}.$
$=1+\mathrm{log}2$, giving the final answer $2e$

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