$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $ Calculate below limit
$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$
Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the resemblence of numerator and denominator?

Using binomial first order expansion we have

*

*$\sqrt{x^3+1}=\sqrt{2+(x^3-1)}=\sqrt 2\sqrt{1+\frac{x^3-1}2}\sim \sqrt 2+\sqrt 2\frac{x^3-1}4$


*$\sqrt{x^4+1}\sim \sqrt 2+\sqrt 2\frac{x^4-1}4$


*$\sqrt{x+1}\sim \sqrt 2+\sqrt 2\frac{x-1}4$


*$\sqrt{x^2+1}\sim \sqrt 2+\sqrt 2\frac{x^2-1}4$
then
$$\frac{x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1}}{x-1 + \sqrt{x+1} - \sqrt{x^2+1}}\sim \frac{x^2-1+\sqrt 2\frac{x^3-1}4-\sqrt 2\frac{x^4-1}4}{x-1+\sqrt 2\frac{x-1}4-\sqrt 2\frac{x^2-1}4}=$$
$$=\frac{x+1+\sqrt 2\frac{x^2+x+1}4-\sqrt 2\frac{x^3+x^2+x+1}4}{1+\sqrt 2\frac{1}4-\sqrt 2\frac{x+1}4}\to \frac{2-\frac14 \sqrt 2}{1-\frac14 \sqrt 2}=\frac{15+2\sqrt2}{7}$$
