Wondering if a limit exists or not

  • #1
mcastillo356
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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##
 

Answers and Replies

  • #2
ergospherical
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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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  • #3
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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.
 
Last edited:
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  • #4
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[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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