Desirae Washington

Answered

2022-07-13

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

Answer & Explanation

Feriheashrz

Expert

2022-07-14Added 8 answers

You have that

${e}^{1+x{2}^{\frac{1}{x}}-x}={e}^{1+x({2}^{\frac{1}{x}}-1)}$

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 ${e}^{1+x({2}^{\frac{1}{x}}-1)}$, putting $z=\frac{1}{x}$ you have that $z\to 0$ as $x\to \mathrm{\infty}$ and so

$\underset{x\to \mathrm{\infty}}{lim}{e}^{1+x({2}^{\frac{1}{x}}-1)}=\underset{z\to {0}^{+}}{lim}{e}^{1+\frac{{2}^{z}-1}{z}}={e}^{1+\mathrm{log}2}=e\cdot {e}^{\mathrm{log}2}=2e$

Alternatively, notice that since when $x\to \mathrm{\infty}$ it is $\frac{1}{x}+{2}^{1/x}>0$, it is

${(\frac{1}{x}+{2}^{1/x})}^{x}=\mathrm{exp}[x\mathrm{log}(\frac{1}{x}+{2}^{1/x})]=\mathrm{exp}\left\{x[\frac{1}{x}\mathrm{log}2+\mathrm{log}(1+\frac{1}{x\cdot {2}^{1/x}})]\right\}$

Now use that $\mathrm{log}(1+t)=t+\text{o}(t)$ as $t\to 0$; I'm using $\mathrm{exp}[f(x)]={e}^{f(x)}$ for better formatting.

${e}^{1+x{2}^{\frac{1}{x}}-x}={e}^{1+x({2}^{\frac{1}{x}}-1)}$

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 ${e}^{1+x({2}^{\frac{1}{x}}-1)}$, putting $z=\frac{1}{x}$ you have that $z\to 0$ as $x\to \mathrm{\infty}$ and so

$\underset{x\to \mathrm{\infty}}{lim}{e}^{1+x({2}^{\frac{1}{x}}-1)}=\underset{z\to {0}^{+}}{lim}{e}^{1+\frac{{2}^{z}-1}{z}}={e}^{1+\mathrm{log}2}=e\cdot {e}^{\mathrm{log}2}=2e$

Alternatively, notice that since when $x\to \mathrm{\infty}$ it is $\frac{1}{x}+{2}^{1/x}>0$, it is

${(\frac{1}{x}+{2}^{1/x})}^{x}=\mathrm{exp}[x\mathrm{log}(\frac{1}{x}+{2}^{1/x})]=\mathrm{exp}\left\{x[\frac{1}{x}\mathrm{log}2+\mathrm{log}(1+\frac{1}{x\cdot {2}^{1/x}})]\right\}$

Now use that $\mathrm{log}(1+t)=t+\text{o}(t)$ as $t\to 0$; I'm using $\mathrm{exp}[f(x)]={e}^{f(x)}$ for better formatting.

Shea Stuart

Expert

2022-07-15Added 4 answers

Taking the logarithm of the function and setting $t:={\displaystyle \frac{1}{x}}$, we can use L'Hospital:

$\underset{t\to 0}{lim}\frac{\mathrm{log}(t+{2}^{t})}{t}=\underset{t\to 0}{lim}\frac{(1+{2}^{t}\mathrm{log}2)}{t+{2}^{t}}.$

$=1+\mathrm{log}2$, giving the final answer $2e$

$\underset{t\to 0}{lim}\frac{\mathrm{log}(t+{2}^{t})}{t}=\underset{t\to 0}{lim}\frac{(1+{2}^{t}\mathrm{log}2)}{t+{2}^{t}}.$

$=1+\mathrm{log}2$, giving the final answer $2e$

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