How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
 A: $$x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)=\frac{x\left(\left(\sqrt{x^2-x}\right)^2-\left(\sqrt{x^2-1}\right)^2\right)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
Since we're searching for the limit as $x\to -\infty$, let $x<0$. Then:
$$=\frac{\frac{1}{-x}(x(1-x))}{\sqrt{\frac{x^2}{(-x)^2}-\frac{x}{(-x)^2}}+\sqrt{\frac{x^2}{(-x)^2}-\frac{1}{(-x)^2}}}=\frac{x-1}{\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}}\stackrel{x\to -\infty}\to -\infty$$
Because $\sqrt{1-\frac{1}{x}}\stackrel{x\to -\infty}\to 1$ and $\sqrt{1-\frac{1}{x^2}}\stackrel{x\to -\infty}\to 1$ and $x-1\stackrel{x\to -\infty}\to -\infty$.
A: 
This is the graph of $x(\sqrt{x^2-x}-\sqrt{x^2-1})$.It is tending to $-\infty$ as $x$ is tending to $-\infty$
After rationalising 
$L=\lim_{x\to -\infty}\frac{x(1-x)}{\sqrt{x^2}(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{x(1-x)}{|x|(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}$
$L=\lim_{x\to -\infty}\frac{x(1-x)}{-x(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{(1-x)}{-(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=-\infty$
Because when $x$ is negative ,$|x|=-x$,not $x$
A: The first step, always, is to get some sound  intuition about the limits. You can draw a graph, but it is machine stuff and can be misleading. You'd better use tools you know, like Taylor series. Here, issues revolve around square rots. Square rots are complicated, except around $1$, where they can be linearized. So, factorize, to get square roots around $1$. You get:
$$ x |x| \left(\sqrt{1-\frac{1}{x}} - \sqrt{1-\frac{1}{x^2}} \right)\,.$$
Now, you start suspecting that the behavior could be of second order under the roots. Using $\sqrt{1-u} \sim 1-\frac{1}{2}u+\frac{1}{8}u^2+O(u^3)$, you can see that $-\frac{1}{x}$ is going to win:
$$\left(\sqrt{1-\frac{1}{x}} - \sqrt{1-\frac{1}{x^2}} \right) \sim -\frac{1}{2x}+\frac{1}{8x^2}-\frac{1}{2x^2}+O\left(\frac{1}{x^3}\right)\,.$$
Your function is likely to  behave, close to $\pm\infty$ as $-\frac{|x|}{2}$ (you thus kill two birds with one stone), hence your limit will be $-\infty$.
Now you can use many different tools to continue trainin your mathematical  skill, like de L'Hospital rule. One idea:  divide by $-\frac{x}{2}$, and prove the product tends to $1$. The more different proofs you find, the more confident you will become in addressing novel problems.
A: You can do the substitution $x=-1/t$ that transforms the limit into
\begin{align}
\lim_{x \to -\infty} x\bigl(\sqrt{x^2-x}-\sqrt{x^2-1}\,\bigr)
&=
\lim_{t\to0^+}-\frac{1}{t}\left(\sqrt{\frac{1}{t^2}+\frac{1}{t}}-\sqrt{\frac{1}{t^2}-1}\,\right)\\[6px]
&=
\lim_{t\to0^+}\frac{\sqrt{1-t^2}-\sqrt{1+t}}{t^2}\\[6px]
&=\lim_{t\to0^+}\frac{1+o(t)-1-\frac{1}{2}t+o(t)}{t}\cdot\frac{1}{t}\\[6px]
&=-\infty
\end{align}

Comments
The substitution $x=-1/t$ will bring the limit into fraction form, usually better manageable than products like your initial form. Why not doing $x=1/t$? Because having a “positive $t$” helps in avoiding mistakes when pulling $t$ outside the square root.
You can avoid the Taylor expansion part by noticing that
$$
\lim_{t\to0^+}\frac{\sqrt{1-t^2}-\sqrt{1+t}}{t}
$$
is the derivative at $0$ of
$$
f(t)=\sqrt{1-t^2}-\sqrt{1+t}
$$
and
$$
f'(t)=-\frac{t}{\sqrt{1-t^2}}-\frac{1}{2\sqrt{1+t}}
$$
so $f'(0)=-\frac{1}{2}$.
A: $$\lim_{x\to -\infty}x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)$$
$$=\lim_{x\to -\infty}x\frac{\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}{\left(\sqrt{x^2-x}+\sqrt{x^2-1}\right)}$$
$$=\lim_{x\to -\infty}x\frac{\left(1-x\right)}{|x|\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$
$$=\lim_{x\to -\infty}\frac{x\left(1-x\right)}{(-x)\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$
$$=-\lim_{x\to -\infty}x\frac{\left(\frac 1x-1\right)}{\left(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}$$
$$=-\lim_{x\to +\infty}x\frac{\left(1+\frac 1x\right)}{\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}\right)}\longrightarrow \color{red}{-\infty}$$
A: Hint: multiply by
$$\frac{\sqrt{x^2-x}+\sqrt{x^2-1}}{\sqrt{x^2-x}+\sqrt{x^2-1}}.$$
You will then probably want to factor an $x^2$ out of the roots in the denominator (note that $\sqrt{x^2} = |x| = -x$ since $x<0$), so you can rewrite the above fraction as
$$\frac{\sqrt{x^2-x}+\sqrt{x^2-1}}{-x\left(\sqrt{1-\frac1x}+\sqrt{1-\frac1{x^2}}\right)}.$$
A: $$\lim _{ x\to -\infty  } \left( \frac { x\left( \sqrt { x^{ 2 }-x } -\sqrt { x^{ 2 }-1 }  \right) \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 }  }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 }  }  \right) =\\ =\lim _{ x\to -\infty  } \left( \frac { x\left( x^{ 2 }-x-x^{ 2 }+1 \right)  }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 }  }  \right) =\lim _{ x\to -\infty  } \left( \frac { x\left( 1-x \right)  }{ \sqrt { x^{ 2 }-x } +\sqrt { x^{ 2 }-1 }  }  \right) =\\ \\ =\lim _{ x\to -\infty  } \frac { x\left( 1-x \right)  }{ \left| x \right| \left( \sqrt { 1-\frac { 1 }{ x }  } +\sqrt { 1-\frac { 1 }{ { x }^{ 2 } }  }  \right)  } =\\ \lim _{ x\to -\infty  } \frac { x-1 }{ \left( \sqrt { 1-\frac { 1 }{ x }  } +\sqrt { 1-\frac { 1 }{ { x }^{ 2 } }  }  \right)  } =-\infty  $$
A: Given
$$\lim_{x\to-\infty}(x(\sqrt{x^2-x}-\sqrt{x^2-1}))$$
As
$$x(\sqrt{x^2-x}-\sqrt{x^2-1}) = \frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{x-x^2}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{\frac{1}{x}-1}{\sqrt{\frac{1}{x^2}-\frac{1}{x^3}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^4}}}$$
$$\lim_{x\to-\infty}\frac{\frac{1}{x}-1}{\sqrt{\frac{1}{x^2}-\frac{1}{x^3}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^4}}}=-\infty$$
As $x\to-\infty$ then $\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3},\frac{1}{x^4}\to0$ And $\frac{-1}{0}=-\infty$
