Find the $\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}}$

Question

How to solve the limits without using L-hospital law, like using rationalisation
L-hospital method is taking too long

The final answer I got was $$\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}} = 1$$

Answer 1

$$\lim_\limits{n\to\infty}{\frac{\sqrt{n^2+1}+\sqrt{n}}{(n^4+1)^{1/4}-\sqrt{n}}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}{((n^4+1)^{1/4}-\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})}{(n^{4}+1)^{\frac{1}{2}}-n}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})((n^{4}+1)^\frac{1}{2}+n)}{((n^{4}+1)^{\frac{1}{2}}-n)((n^{4}+1)^\frac{1}{2}+n)}} = \lim_\limits{n\to\infty}{\frac{(\sqrt{n^2+1}+\sqrt{n})((n^4+1)^{1/4}+\sqrt{n})((n^{4}+1)^\frac{1}{2}+n)}{n^{4}+1 - n^2}}.$$Can you take it from there? Now it is merely a matter of comparing the coefficients of the highest powers of n in the numerator and the denominator.