<s>[INST] Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is grateful. Thanks.
 [/INST] Applying L'Hopital's Rule, we get 
$$
\lim_{x\rightarrow1}{\frac{\frac{1}{2\sqrt{x+1}}+\frac{2x}{2\sqrt{x^2-1}}+\frac{3x^2}{2\sqrt{x^3+1}}}{\frac{1}{2\sqrt{x-1}}+\frac{2x}{2\sqrt{x^2+1}}+\frac{4x^3}{2\sqrt{x^4+1}}}}.
$$
The $2$ in each denominator cancels, and we multiply the numerator and denominator by 
$$
\sqrt{x-1}\cdot\sqrt{x^2+1}\cdot\sqrt{x^4+1}
$$
to get
$$
\lim_{x\rightarrow1}{\frac{\left(\sqrt{x-1}\cdot\sqrt{x^2+1}\cdot\sqrt{x^4+1}\right)\frac{1}{\sqrt{x+1}}+\frac{2x}{\sqrt{x^2-1}}+\frac{3x^2}{\sqrt{x^3+1}}}{\sqrt{x^2+1}\sqrt{x^4+1}+2x\sqrt{x-1}\sqrt{x^4+1}-4x^3\sqrt{x-1}\sqrt{x^2+1}}}.
$$
Substituting $x=1$ in the denominator yields $2$, and distributing in the numerator gives us
$$
\frac{1}{2}\lim_{x\rightarrow 1}{\left( \frac{\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x+1}}+\frac{2x\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x^2-1}}-\frac{3x^2\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x^3+1}} \right)}.
$$
The first and third terms go to zero, so factoring the denominator gives us
$$
\frac{1}{2}\lim_{x\rightarrow 1}{\frac{2x\sqrt{x-1}\sqrt{x^2+1}\sqrt{x^4+1}}{\sqrt{x-1}\sqrt{x+1}}}.
$$
The $\sqrt{x-1}$ terms cancel, and plugging in $x=1$ gives us
$$
\frac{1}{2}\cdot2\sqrt{2}=\sqrt{2}.
$$
This isn't exactly the simplest or most elegant method, but it works.
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