<s>[INST] 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.
 [/INST] 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 $y=0$
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