Find the limit of $x +\sqrt{x^2 + 8x}$ as $x\to-\infty$

Question

$$\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}$$

I multiplied it by the conjugate:

$\frac{-8x}{x - \sqrt{{x^2} + 8x}}$

I can simplify further and get:

$\frac{-8}{1-\sqrt{1+\frac{8}{x}}}$

I think there is an error with my math, because the denominator should probably be a 2.

I'm stuck on this one. I graphed it so I know the limit is -4, but I can't calculate it. Thanks a lot for the help!

Answer 1

We have \begin{align} \lim_{x\to -\infty} x +\sqrt{x^2 + 8x}&=\lim_{x\to -\infty}\frac{(x +\sqrt{x^2 + 8x})(x -\sqrt{x^2 + 8x})}{x -\sqrt{x^2 + 8x}}\\ &=\lim_{x\to -\infty}\frac{x^2 -(x^2 + 8x)}{x -\sqrt{x^2 + 8x}}\\ &=\lim_{x\to -\infty}\frac{-8x}{x -\sqrt{x^2 + 8x}}\\ &=\lim_{x\to -\infty}\frac{\frac{-8x}{|x|}}{\frac{x}{|x|} -\sqrt{\frac{x^2}{|x|^2} + \frac{8x}{|x|^2}}}&& |\cdot |\text{ is needed since }x<0\\ &=\lim_{x\to -\infty}\frac{8}{-1 -\sqrt{1 - 8/x}}&&\text{since }|x|=-x\,\text{ for }x<0\\ &=\frac{8}{-1-\sqrt{1-0}}\\ &=\color{blue}{-4} \end{align}

Answer 2

Notice, $$\lim_{x\to -\infty}(x+\sqrt{x^2+8x})$$ substituting $x=-x$ $$=\lim_{x\to \infty}(-x+\sqrt{x^2-8x})$$ $$=\lim_{x\to \infty}(\sqrt{x^2-8x}-x)\frac{(\sqrt{x^2-8x}+x)}{(\sqrt{x^2-8x}+x)}$$ $$=\lim_{x\to \infty}\frac{x^2-8x-x^2}{\sqrt{x^2-8x}+x}$$ $$=-8\lim_{x\to \infty}\frac{x}{x\sqrt{1-\frac{8}{x}}+x}$$ $$=-8\lim_{x\to \infty}\frac{x}{x\left(\sqrt{1-\frac{8}{x}}+1\right)}$$ $$=-8\lim_{x\to \infty}\frac{1}{\sqrt{1-\frac{8}{x}}+1}$$ $$=-8\left(\frac{1}{\sqrt{1-0}+1}\right)$$ $$=-\frac{8}{2}=\color{red}{-4}$$

Answer 3

The limit as the question stands does not exist. I assume you want to calculate the limit

$\lim_{x\to -\infty} x -\sqrt{x^2 + 8x}$

take $x=-y$

Now, you ca follow you procedure

$\lim_{y\to \infty} -y -\sqrt{y^2 + 8x}$

$=\lim_{y\to \infty}\frac{-8y}{\sqrt{y^2-8y}+y}$

$=\lim_{y\to \infty}\frac{-8}{\sqrt{y-8}+2}$

Now apply the limit, you get $-4$

Answer 4

The problem is with your "simplify further". You are forgetting that $x<0$, and that $\sqrt{x^2}=|x|$. You have, since $|x|/x=-1$, $$ \frac{-8x}{x - \sqrt{{x^2} + 8x}} =\frac{-8}{1 - \frac{\sqrt{{x^2} + 8x}}x} =\frac{-8}{1 - \frac{|x|\sqrt{{1} + \frac8x}}x} =\frac{-8}{1 + \sqrt{{1} + \frac8x}}\longrightarrow\frac{-8}2=-4. $$

Answer 5

Set $-1/x=y, x\to-\infty\implies y\to0^+, y>0$

$\sqrt{x^2+8x}=\sqrt{\dfrac{1-8y}{y^2}}=\dfrac{\sqrt{1-8y}}{|y|}=\dfrac{\sqrt{1-8y}}y$ as $y>0$

$$\lim_{x\to -\infty} x +\sqrt{x^2 + 8x}=\lim_{y\to0^+}\dfrac{-1+\sqrt{1-8y}}y =\lim_{y\to0^+}\dfrac{1-8y-1}{y(\sqrt{1-8y}+1)}=\cdots=\dfrac{-8}{\sqrt1+1}=?$$