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.
 A: Since plugging in $x=1$ gives indeterminate form $\frac{0}{0}$, perhaps try L'Hospital's Rule?
A: I will compute the limits
of the numerator and denominator separately.
To make this more rigorous, imagine that
the two limits are being done
at the same time,
so the final division is justified.
$\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}
=\sqrt{x+1}(1+\sqrt{x-1}-\sqrt{x^2-x+1})
$
so,
putting $x = 1+y$
and using $\sqrt{1+z} \approx 1+z/2$ for small $z$,
$\begin{align}
\lim_{x\to 1} \sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}
&= \lim_{x\to 1}\sqrt{x+1}(1+\sqrt{x-1}-\sqrt{x^2-x+1})\\
&= \lim_{y\to 0}\sqrt{y+2}(1+\sqrt{y}-\sqrt{1+2y+y^2-y-1+1})\\
&= \lim_{y\to 0}\sqrt{2}(1+\sqrt{y}-\sqrt{1+y+y^2})\\
&\approx \sqrt{2}(1+\sqrt{y}-(1+y/2))\\
&\approx \sqrt{2y}
\end{align}
$
Similarly,
$\begin{align}
\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}
&=\sqrt{y}+\sqrt{1+1+2y+y^2}-\sqrt{1+1+4y+6y^2+4y^2+y^4}\\
&=\sqrt{y}+\sqrt{2}(\sqrt{1+y+y^2/2}-\sqrt{1+2y+3y^2+2y^3+y^4/2})\\
&\approx \sqrt{y}+\sqrt{2}((1+y/2)-(1+y))\\
&= \sqrt{y}+\sqrt{2}(-y/2)\\
\end{align}
$
so
$\lim_{x\to 1} \sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}
= \lim_{y\to 0} \sqrt{y}+\sqrt{2}(-y/2)
= \lim_{y\to 0} \sqrt{y}
$.
The ratio of these two is thus
$ \lim_{y\to 0} \frac{ \sqrt{2y}}{\sqrt{y}}
= \sqrt{2}$
A: For every $x>1$ we have
\begin{eqnarray}
P(x):&=&\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}=\sqrt{x^2-1}+\frac{x+1-(x^3+1)}{\sqrt{x+1}+\sqrt{x^3+1}}\\
&=&\sqrt{x^2-1}-\frac{x(x^2-1)}{\sqrt{x+1}+\sqrt{x^3+1}}=\sqrt{x^2-1}\left(1-\frac{x\sqrt{x^2-1}}{\sqrt{x+1}+\sqrt{x^3+1}}\right)\\
&=&\sqrt{x-1}p(x),
\end{eqnarray}
with
$$
p(x):=\sqrt{x+1}\left(1-\frac{x\sqrt{x^2-1}}{\sqrt{x+1}+\sqrt{x^3+1}}\right).
$$
Similarly for every $x>1$ we have
\begin{eqnarray}
Q(x):&=&\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}=\sqrt{x-1}+\frac{x^2+1-(x^4+1)}{\sqrt{x^2+1}+\sqrt{x^4+1}}\\
&=&\sqrt{x-1}+\frac{x^2(1-x^2)}{\sqrt{x^2+1}+\sqrt{x^4+1}}=\sqrt{x-1}\underbrace{\left(1-\frac{x^2(x+1)\sqrt{x-1}}{\sqrt{x^2+1}+\sqrt{x^4+1}}\right)}_{q(x)}.
\end{eqnarray}
It follows that
$$
\lim_{x \to 1^+}\frac{P(x)}{Q(x)}=\lim_{x\to 1^+}\frac{p(x)}{q(x)}=\frac{p(1)}{q(1)}=\sqrt{2}.
$$
A: To ease typing, we temporarily manipulate the top and bottom separately.  
The top is $\sqrt{(x-1)(x+1)}+\left(\sqrt{x+1}-\sqrt{x^3+1}\right)$ (we changed the order). "Rationalize" the part in parentheses. The top  becomes
$$\sqrt{(x-1)(x+1)}-\frac{x(x-1)(x+1)}{\sqrt{x+1}+\sqrt{x^3+1}}.\tag{1}$$
Do the same rationalizing trick with the last two terms of the bottom. We get
$$\sqrt{x-1}-\frac{x^2(x+1)(x-1)}{\sqrt{x^2+1}+\sqrt{x^4+1}}.\tag{2}$$
Divide the expressions (1) and (2) by $\sqrt{x-1}$. We get that our original ratio is equal to
$$  \frac{\sqrt{x+1}-\frac{x(x+1)\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x^3+1}}}{1-\frac{x^2(x+1)\sqrt{x-1}}{\sqrt{x^2+1}+\sqrt{x^4+1}}}                                          .\tag{3}$$
Finally, let $x\to 1^+$. The messy terms in the top and bottom of (3) approach $0$ because of the surviving factor of $\sqrt{x-1}$. So the required limit as $x$ approaches $1$ from the right is $\sqrt{2}$.  
A: I suggest you to make substitution $t = x - 1$. After that you'll get $\lim_{t \rightarrow 0 \dots}$. And next use asymptotic $(1 + x)^p = 1 + px + O(x^2)$ as $x \rightarrow 0$. If necessary you can use more accurate expression by keeping more members in taylor series for $(1+x)^p$.  
A: 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.
