Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$

Question

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.

Answer 1

Notice, $$\lim_{x\to -\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$ $$=\lim_{x\to \infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)$$ $$=\lim_{x\to \infty}\frac{\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$ $$=\lim_{x\to \infty}\frac{4x^2-6-4x^2+x}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$ $$=\lim_{x\to \infty}\frac{x-6}{x\left(\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}\right)}$$ $$=\lim_{x\to \infty}\frac{1-\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}}$$ $$=\frac{1-0}{\sqrt{4-0}+\sqrt{4-0}}=\frac{1}{2+2}=\color{red}{\frac{1}{4}}$$

Answer 2

hint: Use the identity: $\sqrt{A}-\sqrt{B} = \dfrac{A-B}{\sqrt{A}+\sqrt{B}}$, and $\sqrt{4x^2\pm6} = x\sqrt{4\pm\dfrac{6}{x^2}}$

Answer 3

\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}

Answer 4

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 $y=0$