# Correct way to solve limit with square root in denominator

I would like to know how to correctly solve limit as $$x$$ approaches negative infinity for the following expression.

$$\frac{2x+1}{\sqrt{4x^2-2}+1}$$

I would post my attempt at a solution but I apparently can not post pictures and I do not know how to use the mathematical notation very well. But I attempted to solve it like any "power type" limit, trying to dig out the $$x$$ from the numerator and denominator so it can cancels out. Doing that, I arrived at a result of $$1$$, but according to a smart book it should be $$-1$$.

Any help would be much appreciated, thanks.

• hint: $\sqrt{4x^2}=2|x|=-2x$ when $x<0$ – Vasya Mar 20 '19 at 16:40
• @Vasya Of course I completely ignored that. Thanks a lot! – maranovot Mar 20 '19 at 16:54

Hint: Write the denominator as $$|x|\left(\sqrt{4-\frac{2}{x^2}}+\frac{1}{|x|}\right)$$ and the numerator as $$x\left(2+\frac{1}{x}\right)$$ and note that $$\frac{x}{|x|}=-1$$ if $$x<0$$

Hint:

$$\dfrac{2x+1}{\sqrt{4x^2-2}+1}=\dfrac{2x+1}{|x|\sqrt{4-2/x^2}+1}=\dfrac{2x+1}{-x\sqrt{4-2/x^2}+1}$$ $$=\dfrac{2+1/x}{-\sqrt{4-2/x^2}+1/x}$$

Hint:

Divide numerator and denominator by $$x$$.$$\lim_{x\to-\infty}\dfrac{2x+1}{\sqrt{4x^2-2}+1}=\lim_{x\to -\infty}\dfrac{2+1/x}{-\sqrt{4-2/x^2}+1/x}\to-1$$

Aliter:

Let $$t=-x \iff t\to \infty$$ as $$x\to -\infty$$. Therefore the limit translates to: \begin{aligned}\lim_{x\to -\infty}\dfrac{2x+1}{\sqrt{4x^2-2}+1}&=\lim_{t\to \infty}\dfrac{-2t+1}{\sqrt{4t^2-2}+1}\\&=\lim_{t\to\infty}\dfrac{-2+1/t}{\sqrt{4-2/t^2}+1}\to -1\end{aligned}

Hint:

Set $$-\dfrac1x=y\iff x=-\dfrac1y\implies y\to0^+,y>0$$

$$\sqrt{4x^2-2}=\sqrt{\dfrac{4-2y^2}{y^2}}=\dfrac{\sqrt{4-2y^2}}y$$ as $$\sqrt{y^2}=|y|$$ which is $$+y$$ as $$y\ge0$$