B Wondering if a limit exists or not

  • B
  • Thread starter Thread starter mcastillo356
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary
The discussion centers on the limit ##\lim_{x \to{-}\infty}{(\sqrt{x^2+x}-x)}##, with conflicting conclusions about its existence. The textbook states that the limit does not exist because as ##x## approaches negative infinity, the expression grows arbitrarily large. However, online calculators suggest the limit is ##\infty##, leading to confusion about the definition of limits involving infinity. The argument presented indicates that while the expression can increase indefinitely, it does not converge to a specific value, reinforcing the textbook's stance that the limit does not exist. Ultimately, the debate highlights the nuances in understanding limits and the implications of infinity in calculus.
mcastillo356
Gold Member
Messages
641
Reaction score
349
TL;DR
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##
 
Physics news on Phys.org
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##).
 
  • Informative
Likes mcastillo356
##\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.
 
Last edited:
  • Informative
Likes mcastillo356
mcastillo356 said:
[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##.
 
  • Love
Likes mcastillo356

Similar threads