For $g:\ \Bbb{R}\to\Bbb{R}$, it is possible that $$\lim\limits_{x\to -3}\frac{g(x)-g(-3)}{x-(-3)}=5$$ with $\lim\limits_{n\to \infty}g\left(-3+\frac{1}{n}\right)=7$, and $\lim\limits_{n\to \infty}g\left(-3+\frac{\pi}{n^2}\right)=5$.

Okay, so I know the last two limits are essentially the same thing, so the answer must be false. But how do I prove that using the definition of a limit?

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    $\begingroup$ Can you please write it out in LaTeX, cause I can't read the picture: it's too small. Also, are you asking if all of these conditions are true for a single $g$ or for a single $g$? $\endgroup$ May 16, 2015 at 2:04
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    $\begingroup$ In order for $g$ to be differentiable at $-3$, it must be continuous at $-3$... $\endgroup$
    – shalop
    May 16, 2015 at 2:10
  • $\begingroup$ The last two limits show it's not continuous. $\endgroup$ May 16, 2015 at 2:12
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    $\begingroup$ Thank you so much for the help everyone! Also @AlexeyBurdin the book is Elementary Analysis: The Theory of Calculus 2nd Ed. by Ross $\endgroup$ May 16, 2015 at 2:34
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    $\begingroup$ In response to a system flag (caused by a large number of comments) I purged the thread quite a bit. The exchange was about a misundertanding a single user had, and are not entirely relevant to the question at hand. Some quality comments became collateral damage - I apologize for that. If you still need to review the deleted comments, please @-ping me or flag another moderator to undelete the comments. $\endgroup$ May 16, 2015 at 7:23

2 Answers 2


Since $\lim_{x \to -3}\dfrac{g(x)-g(-3)}{x-(-3)} = 5$, we have $$\lim_{x \to -3} \left(g(x) - g(-3)\right) = 5\cdot \lim_{x \to -3} \left(x-(-3)\right) \implies \lim_{x \to -3}g(x) = g(-3)$$ However, $\lim_{n \to \infty} g\left(-3+\dfrac1n\right) = 7$ and $\lim_{n \to \infty} g\left(-3+\dfrac{\pi}{n^2}\right) = 5$, contradicting the fact that $\lim_{x \to -3}g(x) = g(-3)$.

  • $\begingroup$ "$\lim\limits_{x \to -3} \left(g(x) - g(-3)\right)=\dots$" under an assumption that this limit exists. How do I prove the existence? $\endgroup$ May 16, 2015 at 2:30
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    $\begingroup$ @AlexeyBurdin If $\lim_x f(x)$ and $\lim_x g(x)$ exists, then so does $\lim_x f(x)g(x)$. $\endgroup$
    – Adhvaitha
    May 16, 2015 at 2:34
  • $\begingroup$ @AlexeyBurdin He's reached a contradiction by assuming (all) the limits exist. Therefore that cannot be the case (some limit must not exist or not have the values indicated). $\endgroup$
    – Real
    May 16, 2015 at 2:35

Hint: $\displaystyle \lim_{x \to a}f(x) = L$ if and only if, for every sequence $\left \{p_n \right \}$ such that

  • $\displaystyle \lim_{n\to \infty}p_n=a$
  • $p_n \neq a$ for all $n$

Then $$\lim_{n \to \infty}f(p_n)=L$$

  • $\begingroup$ We don't need sequences for this: this is unnecessarily complicated. $\endgroup$ May 16, 2015 at 2:13
  • $\begingroup$ $g(-3+{1 \over n})$ is a sequence $\endgroup$
    – Reveillark
    May 16, 2015 at 2:15

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