Question

Can you solve this without L'Hopital's Rule:
sqrt(x^2 + 8x) -x

Answer 1

you can't solve it with lhopital, it is not a fraction

Answer 2

@CalStu this is very easy to solve it simple just make equal zero ,add to both sides x ,square both sides ,and will get

x^2 +8x = x^2 subtract from both sides x^2 and will get

8x = 0 divide both sides by 8 and will get

x = 0

hope helped

Answer 3

I assume the limit is at either \(\pm \infty\)...
\[\begin{align*}
\lim_{x\to \pm\infty}\left(\sqrt{x^2+8x}-x\right)&=\lim_{x\to \pm\infty}\left(\sqrt{x^2+8x}-x\right)\times\frac{\sqrt{x^2+8x}+x}{\sqrt{x^2+8x}+x}\\[1ex]
&=\lim_{x\to\pm\infty}\frac{\left(\sqrt{x^2+8x}\right)^2-x^2}{\sqrt{x^2+8x}+x}\\[1ex]
&=\lim_{x\to\pm\infty}\frac{x^2+8x-x^2}{\sqrt{x^2+8x}+x}\\[1ex]
&=\lim_{x\to\pm\infty}\frac{8x}{\sqrt{x^2}\sqrt{1+\frac{8}{x}}+x}\\[1ex]
&=\lim_{x\to\pm\infty}\frac{8x}{|x|\sqrt{1+\frac{8}{x}}+x}\\[1ex]
&=\lim_{x\to\pm\infty}\frac{8}{\frac{|x|}{x}\sqrt{1+\frac{8}{x}}+1}
\end{align*}\]The limit then depends on the direction in which \(x\) is heading.

You can use the definition of absolute value to finish the problem:
\[|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x<0\end{cases}\]