Find the limit of $\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}$. Find $$\lim_{n\to \infty}\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}$$
First I tried by taking $\ln y_n=\ln \frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}=\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln(n),$ 
which dose not seems to take me anywhere. Then I tried to use squeeze theorem, since $\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}\geq 1$, but I need an upper bound now. It's for a while I am trying to come up with it but stuck. Can you help me please.
 A: Standard trick when things are difficult due to "mixed" terms like this: Write
$$y_n = \sqrt{n + 1} \ln(n + 1) - \sqrt{n} \ln(n + 1) + \sqrt{n} \ln(n + 1) - \sqrt{n} \ln n$$
Now the first two terms can be combined as
\begin{align*}
\ln(n + 1) \big(\sqrt{n + 1} - \sqrt{n}\big) &= \ln(n + 1) \frac{1}{\sqrt{n + 1} + \sqrt{n}} \\
&\approx \frac{\ln n}{2\sqrt{n}}
\end{align*}
which tends to zero via any number of methods. The last two terms are handled similarly, from which you can deduce that the limit of $y_n$ is zero, and the original limit is $1$.
A: We can do a bit more. Starting with what you wrote $$\ln (y_n)=\ln \frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}=\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln(n)$$ rewrite $$\sqrt{n+1}\ln(n+1)=\sqrt n \sqrt{1+\frac 1n}\Big(\ln(n)+\ln(1+\frac 1n)\Big)$$ and now use Taylor series $$\sqrt{1+\frac 1n}=1+\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\ln(1+\frac 1n)=\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ which makes $$\sqrt{n+1}\ln(n+1)=\sqrt n \Big( 1+\frac{1}{2 n}-\frac{1}{8 n^2}+O\left(\frac{1}{n^3}\right)\Big)\Big(\ln(n)+\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)\Big) $$  Now, replacing $$\ln (y_n)=\sqrt{\frac{1}{n}} \left(1-\frac{1}{2} \ln
   \left(\frac{1}{n}\right)\right)+\frac{1}{8} \left(\frac{1}{n}\right)^{3/2} \ln
   \left(\frac{1}{n}\right)+O\left(\frac{1}{n^{5/2}}\right)$$  Now, using $y_n=e^{\ln(y_n)}$ and Taylor series again $$y_n=1+\sqrt{\frac{1}{n}} \left(1-\frac{1}{2} \log \left(\frac{1}{n}\right)\right)+O\left(\frac{1}{n}\right)$$ which shows the limit and also how it is approached.
For illustartion purposes, let us use $n=10000$ (this is far away from infinity). The rigorous calculation would give $y_n\approx 1.05765$ while the above formula would lead to $\approx 1.05605$.
A: Note that
$$
\begin{align}
1\le\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}
&=\left(1+\frac1n\right)^{\sqrt{n}}(n+1)^{\sqrt{n+1}-\sqrt{n}}\\
&=\left(1+\frac1n\right)^{\large\frac{n}{\sqrt{n}}}
(n+1)^{\large\frac1{\sqrt{n+1}+\sqrt{n}}}\\
&\le e^{\large\frac1{\sqrt{n}}}
\left(\sqrt{n+1}^{\frac1{\sqrt{n+1}}}\right)^2
\end{align}
$$
The Squeeze Theorem then says that
$$
\lim_{n\to\infty}\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}=1
$$
A: We present an approach here that relies only on a standard inequality and the squeeze theorem.  To that end, we proceed.
First, note that we can write the limit of interest as 
$$\lim_{n\to \infty}\left(\frac{(n+1)^{\sqrt{n+1}}}{n^\sqrt{n}}\right)=\lim_{n\to \infty}e^{\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)} \tag 1$$
Next, we analyze the exponent on the right-hand side of $(1)$.  
Using the identity $\log(n+1)=\log(n)+\log\left(1+\frac1n\right)$, the exponent becomes  
$$\begin{align}
\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)&=\left(\sqrt{n+1}-\sqrt{n}\right)\log(n)+\sqrt{n+1}\log\left(1+\frac1n\right)\\\\
&=\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}+\sqrt{n+1}\log\left(1+\frac1n\right) \tag 2\\\\
\end{align}$$

Then, since for any $\alpha>0$, $\log(n)\le \frac{n^\alpha}{\alpha}$, we find that the term of interest satisfies the inequalities

$$0\le\frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}+\sqrt{n+1}\log\left(1+\frac1n\right)\le \frac2{n^{1/4}}+\frac2{\sqrt{n}} \tag 3$$
Applying the squeeze theorem to $(3)$ yields
$$\lim_{n\to \infty} \left(\sqrt{n+1}\log(n+1)-\sqrt{n}\log(n)\right)=0 \tag 4$$
whence using $(4)$ in $(1)$, we find the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{(n+1)^{\sqrt{n+1}}}{n^\sqrt{n}}=1}$$

NOTES:
Note $1$
In this note, we show that the logarithm function satisfies the inequality $\log(n)\le \frac{n^\alpha}{\alpha}$ for any $\alpha >0$.  In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities 
$$\frac{x-1}{x}\le\log(x)\le x-1<x \tag{N1}$$
for $x>0$.  Then, using $\log(n^\alpha)=\alpha \log(n)$ in $(N1)$ with $x=n^\alpha$, we find 
$$\log(n)\le \frac{n^\alpha -1}{\alpha}<\frac{n^\alpha}{\alpha} \tag {N2}$$

Note $2$
In this note, we show that $0\le \frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\le \frac{2}{n^{1/4}}$ for $n\ge 1$.  The left-hand side inequality is trivial.  For the right-hand side inequality we use $(N2)$ with $\alpha =1/4$.  
Then, we see that $\log(n)\le 4n^{1/4}$.  Putting that result together with the inequality $\frac{1}{\sqrt{n+1}+\sqrt{n}}\le \frac{1}{2\sqrt{n}}$, we obtain 
$$0\le \frac{\log(n)}{\sqrt{n+1}+\sqrt{n}}\le \frac{2}{n^{1/4}}$$
as was to be shown.

Note $3$
In this note, we show $0\le \sqrt{n+1}\log\left(1+\frac1n\right)\le \frac2{\sqrt{n}}$, for $n\ge 1$.  The left-hand side inequality is trivial.  For the right-hand side inequality we use $(N1)$ with $x=1+\frac1n$.
Then, we see that $\log\left(1+\frac1n\right)\le \frac1n$.  Putting that result together with $\sqrt{n+1}\le 2\sqrt{n}$, we obtain
$$0\le \sqrt{n+1}\log\left(1+\frac1n\right)\le \frac2{\sqrt{n}}$$ 
as was to be shown. 
A: Notice that for $f(x)=\sqrt x \ln x$ you have
$$f'(x)=\frac{\ln x}{2\sqrt x}-\frac1{\sqrt x}.$$
Now by mean value theorem
$$f(n+1)-f(n) = f'(\theta_n)$$
for some $\theta_n$ such that $n<\theta_n<n+1$ (by Mean Value Theorem). This means that for $n\to\infty$ we have $\theta_n\to\infty$.
If we notice that $\lim\limits_{x\to\infty} f'(x) = 0$ we get that
$$\lim\limits_{n\to\infty} (\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln n) = \lim\limits_{n\to\infty} f'(\theta_n) = 0.$$
