Evaluate $\lim_{x\to \infty}(1+\frac{1}{\sqrt{x}})^x$ 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?
 A: 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 $\infty.$
A: 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 $t\to 0$.
A: $y= (1+\frac{1}{\sqrt{x}})^x$ 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 
$$\lim_{x\to\infty} y=\infty$$
A: 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, 
$$\left(1+\frac1{\sqrt{x}}\right)^{x}\ge e^{\frac{x}{1+\sqrt{x}}}\to \infty\,\,\text{as}\,\,x\to \infty$$
A: 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
A: 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 $\lim_{z\to\infty}e^z=\infty$.
A: 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 $(1+\frac{1}{\sqrt{x}})^x\ge1+\sqrt{x}\to\infty$.
A: It is not hard to show that for $x>1$, $$\left(1+\frac1{\sqrt x}\right)^{\sqrt x}>2$$ and your limit is $\infty$.
