How do I evaluate the limit $\lim _{ x\to -\infty } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 } } $ with a radical? I'm trying to evaluate 

$$\lim _{ x\to -\infty  } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 }  } $$

by rationalizing the denominator, but I am not getting anywhere. Can someone please help me with this?
Thanks
 A: $$\lim_{x\to -\infty}\left(\frac{2x-3}{\sqrt{x^2+7x-2}}\right) = \lim_{x\to -\infty}\left(\frac{2\frac{x}{|x|}-\frac{3}{|x|}}{\sqrt{1+\frac{7}{x}-\frac{2}{x^2}}}\right)=-2$$
A: $$\lim _{ x\to -\infty  } \frac { 2x-3 }{ \sqrt { x^{ 2 }+7x-2 }  } =\lim _{ x\to -\infty  } \frac { x\left( 2-\frac { 3 }{ x }  \right)  }{ \left| x \right| \sqrt { \left( 1+\frac { 7 }{ x } -\frac { 2 }{ { x }^{ 2 } }  \right)  }  } =\\ =\lim _{ x\to -\infty  } \frac { x\left( 2-\frac { 3 }{ x }  \right)  }{ -x\sqrt { 1+\frac { 7 }{ x } -\frac { 2 }{ { x }^{ 2 } }  }  } =-2\\ $$
A: hint
near $-\infty $,
$$\sqrt {x^2+7x-2}=\sqrt {x^2 (1+\frac {7}{x}-\frac {2}{x^2})} $$
$$=\color {red}{-}x\sqrt {1+\frac {7}{x}-\frac {2}{x^2}} $$
A: \begin{align}
\lim_{x\to\infty}\frac{2x-3}{\sqrt{x^2+7x-2}}
&= \lim_{x\to\infty}\frac{x(2-\dfrac3x)}{\sqrt{x^2(1+\dfrac7x-\dfrac{2}{x^2})}} \\
&= \lim_{x\to\infty}\frac{x}{\sqrt{x^2}} \times \lim_{x\to\infty}\frac{2-\dfrac3x}{\sqrt{1+\dfrac7x-\dfrac{2}{x^2}}} \\
&= \lim_{x\to\infty}\frac{x}{\sqrt{x^2}} \times \lim_{x\to\infty}\frac{2-0}{\sqrt{1+0-0}} \\
&= \lim_{x\to\infty}\frac{x}{\sqrt{x^2}} \times 2 \\
&= \lim_{x\to\infty}\frac{2x}{|x|}
\end{align}
it is $2$ as $x\to+\infty$ and $-2$ as $x\to-\infty$.
