Question

Lim
x --> negative infinity

x+ sqrt(x^2+2x)

Answer 1

\[\lim_{x\to-\infty}\left(x+\sqrt{x^2+2x}\right)~~?\]

Answer 2

Yes

Answer 3

\[\begin{align*}\lim_{x\to-\infty}\left(x+\sqrt{x^2+2x}\right)&=\lim_{x\to-\infty}\left(x+\sqrt{x^2+2x}\right)\frac{x-\sqrt{x^2+2x}}{x-\sqrt{x^2+2x}}\\
&=\lim_{x\to-\infty}\frac{x^2-(x^2+2x)}{x-\sqrt{x^2+2x}}\\
&=\lim_{x\to-\infty}\frac{-2x}{x-\sqrt{x^2}\sqrt{1+\frac{2}{x}}}\\
&=\lim_{x\to-\infty}\frac{-2x}{x-|x|\sqrt{1+\frac{2}{x}}}
\end{align*}\]
Since \(x\to-\infty\), you have \(|x|=-x\), so
\[\lim_{x\to-\infty}\frac{-2x}{x-|x|\sqrt{1+\frac{2}{x}}}=\lim_{x\to-\infty}\frac{-2x}{x-(-x)\sqrt{1+\frac{2}{x}}}=\lim_{x\to-\infty}\frac{-2x}{x+x\sqrt{1+\frac{2}{x}}}\]

Answer 4

I an very confused

Answer 5

You start off by multiplying the numerator and denominator by the conjugate. In doing so, you get rid of the square root in the numerator, and the rest is simplified to \(-2x\).

In the denominator, you factor \(\sqrt{x^2}\) from the given expression in the square root.

For all \(x\), \(\sqrt{x^2}=|x|\). Since \(x\) is approaching \(-\infty\), you know that \(x\) is negative. By the definition of the absolute value function, \(|x|=-x\) if \(x<0\).

Answer 6

I cant seem to get the final answer

Answer 7

\[\lim_{x\to-\infty}\frac{-2x}{x+x\sqrt{1+\frac{2}{x}}}=\lim_{x\to-\infty}\frac{-2}{1+\sqrt{1+\frac{2}{x}}}=\cdots\]

Answer 8

-2?

Answer 9

Close, in the denominator you have \(1+\sqrt1\) as \(x\to-\infty\).

Answer 10

-1

Answer 11

yes

Answer 12

Thank you verymuch @SithsAndGiggles. Awesome detailed explanation.
I keep getting 1-1 = 0 in the denomintator. So in order to figure out this question, I have to see that \[\sqrt{}x ^{2}\] is just a positive value?

Answer 13

You have to use the fact that
\[\sqrt{x^2}=|x|=\begin{cases}x&\text{for }x\ge0\\\color{red}{-x}&\color{red}{\text{for }x<0}\end{cases}\]
The red part is what we're particularly interested in, because it helps reduce the expression into something we can work with so that the denominator is nonzero.

Answer 14

Thank you, i get it now

Answer 15

You're welcome