# verifying $\lim\limits_{x \to \infty} (1+\frac{1}{\sqrt{x}})^x$

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, $$\,\,\lim\limits_{t \to 0} \frac{\ln(1+t)}{t}=1$$

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

• 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. Jul 12, 2020 at 11:44

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

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

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

There is another way to compute your limit:

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

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

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

$$=e^{+\infty}=$$

$$=+\infty$$.

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 $$\lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^n =\lim_{s\to\infty}\left(1+\frac1{s}\right)^{s^2}\\ \geq \lim_{s\to\infty}\left(1+\frac1{s}\right)^{ks}=e^k$$

• is it possible to say that $$\lim_{s\to\infty}\left(1+\frac1{s}\right)^{s^2}=e^s$$ Jul 12, 2020 at 18:28
• @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. Jul 12, 2020 at 18:34
• @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. Jul 12, 2020 at 18:38