Find the limit of a fraction So far I have, 
$$
\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{1^2+1}}}{x-1}=\lim_{x\to 1}  \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{2}}}{x-1}
$$
I have no idea how to keep going with this, every way I try I get stuck and can't do anything with it. 
 A: More generally,
consider
$$\lim_{x\to a} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{a^2+1}}}{x-a}.$$
I will use just algebra.
If $x \ne a$,
$\begin{align}
\frac{\frac{x}{\sqrt{x^2+1}} - \frac{a}{\sqrt{a^2+1}}}{x-a}
&=\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\\
&=\left(\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\right)
\left(\frac{x\sqrt{a^2+1}+a\sqrt{x^2+1}}{x\sqrt{a^2+1}+a\sqrt{x^2+1}}\right)\\
&=\frac{x^2(a^2+1)-a^2(x^2+1)}
{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x^2a^2+x^2-a^2x^2-a^2}
{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x^2-a^2}
{(x-a)\sqrt{x^2+1}\sqrt{a^2+1}(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x+a}
{\sqrt{x^2+1}\sqrt{a^2+1}(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
\end{align}
$
Letting $x \to a$,
this becomes
$\frac{2a}
{(a^2+1)(2a\sqrt{a^2+1})}
=\frac{1}
{(a^2+1)^{3/2}}
$.
As a derivative,
$\begin{align}
\left(\frac{x}{\sqrt{x^2+1}}\right)'
&=\frac{\sqrt{x^2+1}-x(1/2)(2x)(x^2+1)^{-1/2}}{x^2+1}\\
&=\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-1/2}}{x^2+1}\\
&=\frac{(x^2+1)-x^2}{(x^2+1)^{3/2}}\\
&=\frac{1}{(x^2+1)^{3/2}}\\
\end{align}
$
which is comforting (and much easier).
Note that the "$1$"
in $x^2+1$ and $a^2+1$
can be any value - 
it is just carried along
and,
if the expression is
$\frac{x}{\sqrt{x^2+b}}$,
the result is
$\frac{b}{(x^2+b)^{3/2}}$.
A: Set $f(x)=\frac{x}{\sqrt{x^2+1}}$, $f'(x) = \frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{x^2+1}}}{x^2+1}$ by the quotient rule (I hope, I'm terrible at math).
The answer to your question is $f'(1)$ which shouldn't be hard to calculate.
A: Hint: Use L'Hopital's Rule. Take the derivatives of both the numerator and denominator, then substitute the limit value.
A: Just as for aaa, L'Hopital's rule is, at least to me too, the simplest way to solve your problem. There is another one (which is usable if you know about Taylor series). Around x = 1, the first terms of the Taylor series of x / Sqrt[x^2 + 1] is 1 / Sqrt[2] + (x -1) / (2 Sqrt[2]) +... Then, replace the numerator by this expansion and simplify.
