2
$\begingroup$

Evaluate $\lim_{x\to \infty}(1+\frac{1}{\sqrt{x}})^x$

This is an exercise in my analysis book and I'm having trouble solving it. It's clearly related to the limit of $e$. I've tried rewriting:

Evaluate $\lim_{x\to \infty}((1+\frac{1}{\sqrt{x}})^\sqrt{x})^\sqrt{x}$.

Now I sadly cannot rewrite this as $\lim_{x\to \infty}(e)^x$ or anything similar. Thoughts?

$\endgroup$
  • $\begingroup$ Why can't? It seems to me it approaches infinity. (take log if nessecary). $\endgroup$ – kolobokish Dec 6 '16 at 19:12
  • $\begingroup$ Indeed you cannot. $\endgroup$ – Did Dec 6 '16 at 23:58
4
$\begingroup$

Suppose $x\in [n,n+1),$ where $n$ is a positive integer. Using the binomial theorem we have

$$(1+1/\sqrt x)^x > (1+1/\sqrt {n+1})^n = 1^n + n\cdot 1^{n-1}\cdot (1/\sqrt {n+1})^1 + \cdots > n/\sqrt {n+1}.$$

Since $n/\sqrt {n+1} \to \infty,$ the desired limit is $\infty.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Clean proof. Thanks! $\endgroup$ – QuestionMaker Dec 6 '16 at 19:29
2
$\begingroup$

Hint. Note that as $x\to +\infty$, $$\left(1+\frac{1}{\sqrt{x}}\right)^x=\exp\left(x\ln\left(1+\frac{1}{\sqrt{x}}\right)\right)\sim \exp\left(\frac{x}{\sqrt{x}}\right)=\exp\left(\sqrt{x}\right)$$ wher we use the fact that $\ln(1+t)\sim t$ as $t\to 0$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

$y= (1+\frac{1}{\sqrt{x}})^x$ then $\ln y=\ln(1+\frac{1}{\sqrt{x}})^x=x\ln(1+\frac{1}{\sqrt{x}})$. We have $$\lim_{x\to\infty} x\ln(1+\frac{1}{\sqrt{x}})=\lim_{x\to\infty}=\frac{\ln (1+\frac{1}{\sqrt{x}})}{\frac{1}{x}}=^H\lim_{x\to\infty}\frac{ \frac{\frac{-1}{2x\sqrt{x}}}{\frac{\sqrt{x}+1}{\sqrt{x}}}}{\frac{-1}{x^2}}=\lim_{x\to\infty}\frac{ \frac{-1}{2x(\sqrt{x}+1)}}{\frac{-1}{x^2}}=\infty$$ So $\lim_{x\to\infty} \ln y=\infty$ and consequently $$\lim_{x\to\infty} y=\infty$$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Note that

$$\lim_{x\to \infty}\left(1+\frac1{\sqrt{x}}\right)^{\sqrt{x}}=e $$

Therefore, for sufficiently large $x$, $\left(1+\frac1{\sqrt{x}}\right)^{\sqrt{x}}\ge 2$ (i.e., take $\epsilon = e-2>0$).

Hence, for sufficiently large $x$, we have

$$\left(1+\frac1{\sqrt{x}}\right)^{x}\ge 2^{\sqrt{x}}\to \infty\,\,\text{as}\,\,x\to \infty$$

And we are done!


Alternatively, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm satisfies the inequality

$$\log(x)\ge \frac{x}{1+x}$$

Therefore,

$$\left(1+\frac1{\sqrt{x}}\right)^{x}\ge e^{\frac{x}{1+\sqrt{x}}}\to \infty\,\,\text{as}\,\,x\to \infty$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Evaluate $\lim_{x\to \infty}((1+\frac{1}{\sqrt{x}})^\sqrt{x})^\sqrt{x}$.

Well you know that this simplifies to: $$e^{\sqrt{x}} = e^{\sqrt{\infty}}$$

Square root of infinity is just infinity really, and any number other then 1 raised to infinity is infinity. If you think of infinity as a really large number ... then its just 2.7^(a large number), and thus this will approach infinity

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Compute first the limit of the (natural) logarithm: $$ \lim_{x\to\infty}\log\left(\left(1+\frac{1}{\sqrt{x}}\right)^{\!x}\right)= \lim_{x\to\infty}x\log\left(1+\frac{1}{\sqrt{x}}\right)= \lim_{t\to0^+}\frac{\log(1+t)}{t^2} $$ after the substitution $\sqrt{x}=1/t$, with $t>0$. Now $$ \lim_{t\to0^+}\frac{\log(1+t)}{t}=1 $$ so $$ \lim_{t\to0^+}\frac{\log(1+t)}{t^2}=\infty $$ Therefore also your limit is $\infty$, because $\lim_{z\to\infty}e^z=\infty$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

For $\alpha\ge1, t\ge-1$, $(1+t)^\alpha\ge1+\alpha t$

We can see this by considering the tangent at the origin and convexity. (This is Bernoulli's inequality.)

Take $t=\frac{1}{\sqrt{x}},\alpha=x$ to see $(1+\frac{1}{\sqrt{x}})^x\ge1+\sqrt{x}\to\infty$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

It is not hard to show that for $x>1$, $$\left(1+\frac1{\sqrt x}\right)^{\sqrt x}>2$$ and your limit is $\infty$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.