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

Question

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

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

Answer 1

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

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

Answer 2

There is another way to compute your limit:

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

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

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

$=e^{+\infty}=$

$=+\infty$.

Answer 3

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