Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Solving without L'Hopital
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
That's
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}\right)-\lim_{x\to-\infty}\left(\sqrt{4x^2+x}\right)$$
I have been taught to get the highest exponents, so...
$$\sqrt{4x^2}-\sqrt{4x^2}$$
It's the same for both sides, no?
$$2x-2x$$
$$-\infty+\infty$$
Which is wrong. The correct answer is
$$\frac{1}{4}$$
Why? I always just grab the highest exponent ($\sqrt{4x^2}$) and work with it. But this time it didn't go well.
 A: hint: Use the identity: $\sqrt{A}-\sqrt{B} = \dfrac{A-B}{\sqrt{A}+\sqrt{B}}$, and $\sqrt{4x^2\pm6} = x\sqrt{4\pm\dfrac{6}{x^2}}$
A: \begin{align}
\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)&=\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)\cdot\frac{\sqrt{4x^2-6}+\sqrt{4x^2+x}}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\
&=\lim_{x\to-\infty}\frac{(\sqrt{4x^2-6})^2-(\sqrt{4x^2+x})^2}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\
&=\lim_{x\to-\infty}\frac{4x^2-6-(4x^2+x)}{\sqrt{4x^2-6}+\sqrt{4x^2+x}}\\
&=\lim_{x\to-\infty}\frac{(-x-6)/|x|}{\sqrt{4-6/x^2}+\sqrt{4+x/x^2}}\\
&=\lim_{x\to-\infty}\frac{1+\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}\\
&=\frac{1}{\sqrt{4}+\sqrt{4}}\\
&=\frac{1}{4}
\end{align}
A: Notice, $$\lim_{x\to -\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
$$=\lim_{x\to \infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)$$
$$=\lim_{x\to \infty}\frac{\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$
$$=\lim_{x\to \infty}\frac{4x^2-6-4x^2+x}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$
$$=\lim_{x\to \infty}\frac{x-6}{x\left(\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}\right)}$$
$$=\lim_{x\to \infty}\frac{1-\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}}$$
$$=\frac{1-0}{\sqrt{4-0}+\sqrt{4-0}}=\frac{1}{2+2}=\color{red}{\frac{1}{4}}$$
A: Set $-1/x=y\implies y\to0^+, |y|=+y$ 
and $4x^2-6=\dfrac{4-6y^2}{y^2},\sqrt{4x^2-6}=\dfrac{\sqrt{4-6y^2}}{|y|}=\dfrac{\sqrt{4-6y^2}}y$
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)=\lim_{y\to0^+}\dfrac{\sqrt{4-6y^2}-\sqrt{4-y}}y$$
$$=\lim_{y\to0^+}\dfrac{(4-6y^2)-(4-y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$
$$=\lim_{y\to0^+}\dfrac{y(1-6y)}{y(\sqrt{4-6y^2}+\sqrt{4-y})}$$
Cancel out $y$ as $y\ne0$ as $y\to0$
Then set $y=0$
