The value of $\underset{x\to \mathrm{\infty }}{lim} \frac{\sqrt{n^2+1}+\sqrt{n}}{\sqrt[4]{n^4+n}+\sqrt[4]{n}}$, is 

Solution:

we have,

$\begin{array}{l}\underset{n\to \mathrm{\infty }}{lim} \frac{\sqrt{n^2+1}+\sqrt{n}}{\sqrt[4]{n^3+n-\sqrt[4]{n}}}\\ =\underset{n\to \mathrm{\infty }}{lim} \frac{n\left\{\sqrt{1+\frac{1}{n^2}}+\frac{1}{\sqrt{n}}\right\}}{n^{3/4}\left\{\sqrt[4]{1+\frac{1}{n^2}}-\frac{1}{\sqrt{n}}\right\}}\\ =\underset{n\to \mathrm{\infty }}{lim} n^{1/4}\frac{\left\{\sqrt{1+\frac{1}{n^2}}+\frac{1}{\sqrt{n}}\right\}}{\left\{\sqrt[4]{1+\frac{1}{n^2}}-\frac{1}{\sqrt{n}}\right\}}=\mathrm{\infty }\times 2=\mathrm{\infty }\end{array}$