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A related question to this question, I am wondering $$\lim_{x\to a}{f^\prime(x)}=+\infty,$$ what can be concluded about $f(a)$? Does this invalidate that $f(x)$ is not continuous at $a$ because of the non-existence of $f^\prime(a)$? Does this condition also imply maybe $$\lim_{x\to a}{f(x)}=+\infty?$$

I would think $$\lim_{x\to a}{f(x)}=+\infty,$$ because here $f(a+h)-f(a)$ can be arbitrarily large no matter how small $h$ is.

EDIT

Okay, I see where I got it wrong. Even though $$\lim_{x\to a}{f^\prime(x)}=+\infty,$$ it does not mean $f(a+h)-f(a)$ is arbitrarily large, because an $\infty$ times an infinitesimal quantity may not be determinate.

I just wonder another related question: given $$\lim_{x\to a}{f(x)}=+\infty,$$ what can be concluded to $f^\prime(a)$? Can it be finite or non-existent? How about also when $a=\infty$ in this case.

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2 Answers 2

up vote 9 down vote accepted

Consider $f(x)=\sqrt[3]{x}$, $a=0$. The derivative goes to infinity, but the function is continuous. The graph has a vertical tangent line at $(0,0)$.


For your second question, if $\displaystyle{\lim_{x\to a}f(x)=+\infty}$, then $f(a)$ cannot be defined in a way such that $f$ continuous at $a$, and therefore the derivative at $a$ cannot exist. You can conclude that the derivative near $a$, if it exists, would be unbounded above to the left of $a$ and unbounded below to the right of $a$, by the mean value theorem.

If $\displaystyle{\lim_{x\to +\infty}f(x)=+\infty}$, then it is possible that $\displaystyle{\lim_{x\to \infty}f'(x)}$ exists and is finite. E.g., $f(x)=x$, or $\sqrt[3]{x}$ again. But it could do lots of things. You can be sure that if $f'(x)$ exists everywhere, then the set of $x$ where $f'(x)>0$ is unbounded, by the mean value theorem.

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No, $f(x)$ can still be continuous. Think of $f(x)=1-\sqrt{|x|}$. The derivative limits to $+\infty$ from the left (and $-\infty$ from the right) at $x=0$, but the function is still continuous. It just has a cusp at $0$.

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Jonas beat me to it. –  Grumpy Parsnip Apr 30 '11 at 1:38

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