B Wondering if a limit exists or not

[mcastillo356]

TL;DR Summary

My textbook says the limit does not exist. I don't agree, or there is something I miss.

I have opposite conclusions about $\lim_{x \to{-}\infty}{(\sqrt{x^2+x}-x)}$

Quote from my textbook:
"The limit $\lim_{x \to{-}\infty}{(\sqrt{x^2+x}-x)}$ is not nearly so subtle. Since $-x>0$ as $x\rightarrow{-\infty}$, we have $\sqrt{x^2+x}-x>\sqrt{x^2+x}$, which grows arbitrarily large as $x\rightarrow{-\infty}$. The limit does not exist."

But online limits calculators say the limit is $\infty$, and my personal opinion is:
$\lim_{x \to{-}\infty}{(\sqrt{x^2+x}-x)}=\infty-(-\infty)=\infty$

 

[ergospherical]

For $x < 0$,\begin{align*}
\lim_{x \rightarrow -\infty} \sqrt{x^2 + x} - x &= \lim_{x \rightarrow -\infty} |x| \left(\sqrt{1+\dfrac{1}{x}} + 1 \right) \\
&= \lim_{x \rightarrow -\infty} |x| \left( 2 + \dfrac{1}{2x} + O\left( \dfrac{1}{x^2} \right) \right) \\
&= \lim_{x \rightarrow -\infty} 2|x| - \dfrac{1}{2}
\end{align*}which does not exist (i.e. you can make the thing on the right arbitrarily large for sufficiently negative $x$).

 

[fresh_42]

$\pm \infty$ are not numbers. A sequence is divergent if it increases or decreases forever to $\pm \infty$. It is not convergent to infinity. In this sense, the limit does not exist.

 

[Mark44]

[My textbook says the limit does not exist. I don't agree, or there is something I miss.

It's sort of a semantics thing. Although we can write $\lim_{x \to \infty}x^2 = \infty$, if the "limit" is $\infty$, we consider the limit to not exist, since $\infty$ is not a number in the real number system.

Another way that the limit can fail to exist is if you get different values on either side of the limit point, as in $\lim_{x \to 0} \frac 1 x$.