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If $f$ is an even function such that $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$ has a finite non zero value , then is $f(x)$ continuous , differentiable , or neither continuous nor differentiable at $x=0$?

I think that the function is continuous , and it is easy to prove that too. But it would not be differentiable. This is because , the above expression represents the derivative of the function at $x=0$ and and the derivative of an even function at $0$ is always zero however according to the given question , the derivative of the function is not zero. So it is not differentiable. Do you think I have reasoned it correctly ?

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Xander Henderson
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Aditi
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    The limit you wrote is the definition of the derivative. Do you really want that one or a lateral one like $x\to0^+$? –  Jul 13 '18 at 18:27
  • No according to the given question it’s just $x \to 0$ – Aditi Jul 13 '18 at 18:28
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    You say the derivative of an even function at $0$ is always zero; to make this true, you should add the phrase: "if it exists". Many even functions are not differentiable at $x=0$ – G Tony Jacobs Jul 13 '18 at 18:28
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    If the function is even, and that two-sided limit is non-zero, then we have a contradiction. :/ – G Tony Jacobs Jul 13 '18 at 18:29
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    @Aditi Then everything is true, because on one hand you know that limit exists and it is non-zero, on the other you know that for an even function if that limit exists it should be zero. –  Jul 13 '18 at 18:29
  • But if he derivative exists , it still wouldn’t be differentiable by the reason I’ve stated right ? Or would there be another reason for it’s non differentiability? – Aditi Jul 13 '18 at 18:30
  • @scentofthetrees okay ! That’s what I wanted to know! Thanks ! – Aditi Jul 13 '18 at 18:30
  • @Aditi Take into account that by "everything is true" I really mean everything. $f$ is differentiable, $f$ is not differentiable, $f$ is continuous, $f$ is not continuous, ... –  Jul 13 '18 at 18:32
  • @scentofthetrees the above question was a multiple choice question which had three choices : continuous and differentiable , discontinuous and not differentiable , or just discontinuous – Aditi Jul 13 '18 at 18:34
  • @Aditi If that's the case, then it's not a valid question. Do you still have a copy of the question available? Could you post the question exactly (ideally, both a transcript and a photo of the question)? – Tanner Swett Jul 13 '18 at 18:38
  • @Tanner Okay ! Well just to add a detail, the question had four choices of which the last one was none of these . I thought it was redundant earlier since the answer to the question was continuous but non differentiable but just wanted to check if I’d reasoned it correctly – Aditi Jul 13 '18 at 18:40
  • If the answer is continuous, but not differentiable, then I am convinced that the information given was should have been only the lateral limit. –  Jul 13 '18 at 18:43
  • @scentofthetrees have a look at the original question please – Aditi Jul 13 '18 at 18:44
  • @Aditi I did. For the question as written, all (a),(b),(c),(d) are true. There is clearly a typo, and the given limit should just be a lateral limit, in which case the answer is (b). –  Jul 13 '18 at 18:45
  • Given that the authors of the book cannot decide if a function is "derivable" or "differentiable" at $x=0$, it is no wonder that there are other, more important, typos. – Xander Henderson Jul 13 '18 at 18:46
  • @scentofthetrees alright thank you for your time :) – Aditi Jul 13 '18 at 18:48
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    It might be helpful if we knew which book this is from. It might also be useful to send an email to the author(s) and point out the error (or see if there are errata which correct the error). – Xander Henderson Jul 13 '18 at 19:09
  • @Xander I do know this book . Would you like to know it too ? – Aditi Jul 13 '18 at 19:10
  • I wouldn't have asked if I didn't want an answer. – Xander Henderson Jul 13 '18 at 19:11
  • @Xander I’m sorry. The book is the calculus edition of Cengage publications . It’s a book for high school students aiming for entrance examinations in India . – Aditi Jul 13 '18 at 19:12
  • This problem appears to be from page 12.8 of The Pearson Guide To Complete Mathematics For The Aieee, 4/E. I cannot find any errata online. – Xander Henderson Jul 13 '18 at 19:20
  • @Xander quite true. The general trend in these books is that they circulate almost the same question over again in different books so it’s hard to decipher who had originally formulated it. This specific question was from a daily practice problems book of cengage publications and I think it would be hard to find that specific book online. Thanks a lot for your help and time – Aditi Jul 13 '18 at 19:24

3 Answers3

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By the Principle of Explosion, all answers are correct. There are two contradictory statements in the question:

  • It is stated that $f'(0)$ exists (because the limit of the difference quotient exists) and that $f$ is even. Hence \begin{align} f'(0) = \lim_{h\to 0} \frac{f(h)-f(0)}{h} &= \lim_{h\to 0} \frac{f(-h)-f(0)}{h} \\ &= \lim_{-h\to 0} \frac{f(h)-f(0)}{-h} \\ &= -\lim_{h\to 0} \frac{f(h)-f(0)}{h} \\ &= -f'(0). \end{align} Since $f'(0) = -f'(0)$, it must be the case that $f'(0) = 0$.
  • It is also state that $f'(0)$ exists and $f'(0) > 0$.

These two statements are contradictory, therefore any conclusion follows. The correct answer, then, is to mark all of the multiple choice options. ;)

While it is impossible to know, it is likely that (as has been pointed out in the comments) the author(s) intended to consider the one sided limit. It is reasonable to conclude that the question should have read

If $f$ is an even function such that $$ \lim_{h\to 0^+} \frac{f(h)-f(0)}{h} $$ has some finite, non-zero value, then... (multiple choice options).

In this case, there is no contradiction. Then, by the reasoning above, we know that the function cannot be differentiable at $x=0$. If it were, then the derivative would be zero, but we know that it is not.

On the other hand, suppose that the one-sided limit is $L$, i.e. that $\lim_{h\to 0^+} (f(h)-f(0))/h = L$. Then $$ |f(h) - f(0)| = \left| h \frac{f(h) - f(0)}{h} \right| = |h| \left| \frac{f(h) - f(0)}{h} \right| = |h| |L|. $$ We can make this as small as we like by choosing $h$ sufficiently small. The limit from the left will be the same (the sign of the limits will differ, but this is absorbed into the absolute value). This implies that $f$ is continuous at zero. Therefore, assuming that the question was supposed to read as indicated above, the correct answer would be (b).

Xander Henderson
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  • Thank you ! But do you think it is correct to not consider the lateral limit and only conclude it on the basis of the derivative tending to zero , and the derivative at zero, i.e. , the function does not have a derivative because it’s derivative tending to zero and that at zero is different? – Aditi Jul 13 '18 at 18:47
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    Unless otherwise stated, a limit is two-sided. If the author(s) had wanted you to consider a one-sided limit, then they should have indicated so. – Xander Henderson Jul 13 '18 at 18:48
  • Alright ! Thanks for your help ! – Aditi Jul 13 '18 at 18:50
  • Thanks for the edit ! I’d earlier thought about it too but I was confused as well because the author hadn’t explicitly specified a one sided limit – Aditi Jul 13 '18 at 19:21
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Well, the given question isn't really valid, because it's not possible for $f$ to be an even function such that $\lim_{h \to 0} \frac{f(h)-f(0)}{h}$ has a finite non zero value.

Your reasoning isn't really correct, because you should have concluded that the given conditions are impossible. "The function is not differentiable at $0$" is not the correct conclusion to come to.

In my opinion, none of the given answer choices are accurate.

Tanner Swett
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So I think the answer is that the you are right: function is continuous but not differentiable.

I am assuming that the question meant 'limit from the positive direction' otherwise, as others have pointed out, the question is contradictory. However this is a reasonable assumption because with this assumption the question becomes meaningful and interesting.

The limit 'from the right; the positive side' is non-zero. The function is even and so the limit that comes from the left (the negative side) will have the opposite sign. And so these two limits cannot be equal (because they are non-zero) which means that the function does not have a derivative. For a mental picture, imagine two straight lines (one from the top right and one from the top left) into the origin and forming a point. At the origin there is no derivative.

But the function is continuous essentially because we can make the distance between $f(h)$ and $f(0)$ as small we we please by making $h$ small and then the difference between $f(h)$ and $f(0)$ will look like $h.l$ where $l$ is the limit of $(f(h)-f(0))/h$. We will be able to do something similar (with reversed sign) from the other side because $f$ is even. So, yes, I think you are right: $f$ is continuous but not differentiable.

Simon Terrington
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  • Hi ! I did know the answer to the question already . However I had not considered two sided limits while evaluating it. I had just considered the derivative that was tending to zero and the derivative at zero. As these two were not equal , I concluded that the function is not differentiable. I just wanted to know whether it was correct to not consider two sided limits – Aditi Jul 13 '18 at 18:56
  • Aha, that makes sense :) – Simon Terrington Jul 13 '18 at 19:01
  • Thanks for you help anyways ! – Aditi Jul 13 '18 at 19:02
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    It was good to speak - interesting problem! – Simon Terrington Jul 13 '18 at 19:04
  • May I suggest that you edit your answer to move the "PS" to the top and make it clear from the start that you are making this assumption? – Xander Henderson Jul 13 '18 at 19:32