Find the limit of $\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}$.

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 $$\lim\limits_{n\to\infty} (\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln n) = \lim\limits_{n\to\infty} f'(\theta_n) = 0.$$

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 $1$.

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 $\approx 1.05605$.