What is the limit of $\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x$ ?

I tried to solve it but I am not sure if it is appropriate to solve it this way.

$(1+\frac{1}{\sqrt{x}})^x =\exp\Bigl({x\times\ln\bigl(1+\frac{1}{\sqrt{x}}\bigr)\Bigr)}\tag{*}$

Let $\,t=\frac {1}{\sqrt{x}} $

$\,t=\frac {1}{\sqrt{x}} \Rightarrow t^2=\frac {1}{x}\Rightarrow \frac {1}{t^2}=x $

By substituting in $(*) $ we have

$\exp\Bigl(x\ln\bigl(1+\frac{1}{\sqrt{x}}\bigr)\Bigr)=\exp\Bigl(\frac{1}{t^2}\ln(1+t)\Bigr) $

as $\quad x\rightarrow \infty ,\quad \frac{1}{\sqrt{x}}\rightarrow 0,\quad$ so $t\rightarrow 0 $

$\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x = \lim\limits_{t \to 0} \exp\bigl(\frac{1}{t^2}\ln(1+t)\bigr)$

as $t\neq 0 \,$ we can divide and multiply by $t$: \begin{align} \lim_{t \to 0} \exp\Bigl(\frac{1}{t^2}\times \ln(1+t)\Bigr)&=\lim_{t \to 0} \exp\Bigl(\frac{1}{t^2}\times \ln(1+t)\times \frac{t}{t}\Bigr)\\ &=\lim_{t \to 0} \exp\Bigl(\frac{1}{t}\times \frac {\ln(1+t)}{t}\Bigr) \end{align}

using L’Hospital’s rule, $\,\,\lim\limits_{t \to 0} \frac{\ln(1+t)}{t}=1$

$\lim\limits_{t \to 0}\exp\Bigl(\frac{1}{t}\times \frac {\ln(1+t)}{t}\Bigr)=\infty$

  • 3
    $\begingroup$ Your formatting could use a little bit of work. Here are a few pointers: There is a \exp command. Double dollar signs ($$...$$) rather than single would make much of this look better. And when you have fractions and / or exponents (or similar tall things) inside parentheses, changing (...) to \left(...\right) makes them auto-adjust size. This also works with [...], \{...\} and \lceil...\rceil as well as many other bracket-type symbols and commands. $\endgroup$
    – Arthur
    Jul 12, 2020 at 11:44

3 Answers 3


You can show the limit much quicker by bounding your expression from below using the Bernoulli inequality :

$$\left(1+\frac{1}{\sqrt{x}}\right)^x\geq 1+x\cdot\frac{1}{\sqrt{x}} =1+\sqrt x\stackrel{x\to +\infty}{\longrightarrow}+\infty$$

  • $\begingroup$ (+1) I was just thinking of adding a Bernoulli answer when I saw this. This is definitely simple. $\endgroup$
    – robjohn
    Jul 12, 2020 at 14:07

There is another way to compute your limit:

$\lim\limits_{x \to +\infty} \left(1+\frac{1}{\sqrt{x}}\right)^x=$

$=\lim\limits_{x \to +\infty} \left[\left(1+\frac{1}{\sqrt{x}}\right)^\sqrt{x}\right]^\sqrt{x}=$

$=\left[\lim\limits_{x \to +\infty} \left(1+\frac{1}{\sqrt{x}}\right)^\sqrt{x}\right]^{\lim\limits_{x \to +\infty} \sqrt{x}}=$




I think you ought to specify $\lim_{t\to 0^+}$, just to be on the safe side. Apart from that, it looks correct.

It's a lot quicker, however, to use $s^2=x$ and note that for any real number $k$, we have $$ \lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^n =\lim_{s\to\infty}\left(1+\frac1{s}\right)^{s^2}\\ \geq \lim_{s\to\infty}\left(1+\frac1{s}\right)^{ks}=e^k $$

  • $\begingroup$ is it possible to say that $$ \lim_{s\to\infty}\left(1+\frac1{s}\right)^{s^2}=e^s $$ $\endgroup$
    – Ada Az
    Jul 12, 2020 at 18:28
  • 1
    $\begingroup$ @AdaAz No, it isn't. Your left-hand side is a concrete "number" (it's infinite, but still) that doesn't depend on anything, while your right-hand side depends on the value of the variable $s$, and is finite. Thus there is no way they can be equal. $\endgroup$
    – Arthur
    Jul 12, 2020 at 18:34
  • 1
    $\begingroup$ @AdaAz You can, however, say that as long as $s\geq1$ we have $$2^s\leq\left(1+\frac1s\right)^{s^2}\leq e^s$$and $\leq$ is preserved when taking limits. $\endgroup$
    – Arthur
    Jul 12, 2020 at 18:38

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