Wondering if a limit exists or not

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Discussion Overview

The discussion centers around the limit ##\lim_{x \to{-}\infty}{(\sqrt{x^2+x}-x)}##, exploring whether it exists or not. Participants examine various interpretations and mathematical approaches to the limit, including differing opinions on the implications of reaching infinity.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant cites a textbook stating that the limit does not exist, arguing that as ##x \to -\infty##, the expression grows arbitrarily large.
  • Another participant suggests that online calculators indicate the limit is ##\infty## and expresses a personal belief that the limit can be interpreted as ##\infty##.
  • A mathematical derivation is presented showing that for ##x < 0##, the limit leads to an expression that does not exist, implying it can become arbitrarily large.
  • One participant argues that ##\pm \infty## are not considered numbers and states that a limit is defined as divergent if it approaches ##\pm \infty##, thus concluding the limit does not exist.
  • Another participant reflects on the semantics of limits approaching infinity, suggesting that while limits can be expressed as approaching infinity, they may still be considered as not existing in the context of real numbers.

Areas of Agreement / Disagreement

Participants express differing views on whether the limit exists, with some arguing it does not exist and others suggesting it can be interpreted as ##\infty##. The discussion remains unresolved with multiple competing interpretations.

Contextual Notes

Participants highlight the semantic differences in defining limits that approach infinity and the implications of divergent sequences, indicating a lack of consensus on the definitions and interpretations involved.

mcastillo356
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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##
 
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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##).
 
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##\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.
 
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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##.
 
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