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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.

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

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Hint:

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

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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: $$\begin{aligned}\lim_{x\to -\infty}\dfrac{2x+1}{\sqrt{4x^2-2}+1}&=\lim_{t\to \infty}\dfrac{-2t+1}{\sqrt{4t^2-2}+1}\\&=\lim_{t\to\infty}\dfrac{-2+1/t}{\sqrt{4-2/t^2}+1}\to -1\end{aligned}$$

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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 $y\ge0$

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