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antibody

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Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?

any one can give an example? thx!

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- #1

antibody

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Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?

any one can give an example? thx!

- #2

JustinLevy

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The limit of f'(x) as x-> -1 is 0, however f(-1) is undefined, and therefore I believe f'(-1) is undefined as well.

I'm no mathematician, so can someone comment on whether this reasoning is correct?

- #3

antibody

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The limit of f'(x) as x-> -1 is 0, however f(-1) is undefined, and therefore I believe f'(-1) is undefined as well.

I'm no mathematician, so can someone comment on whether this reasoning is correct?

i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..

- #4

JustinLevy

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i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..

Like I said, it is best to have someone else verify my claim, but I believe the limit does exist. Approach from either side and f'(x) as x-> -1 is just 0. So why do you feel the limit does not exist? f'(-1) may be undefined, but I believe the limit as x-> -1 is.

Can someone verify this reasoning?

- #5

StatusX

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I think the OP is asking if there's an example where both the limit as f'(x) approaches a and f'(a) exist, but are not equal. I don't think this is possible, but I'm having trouble thinking of a simple proof. For example, if you could justify switching the limits, it'd just be:

[tex] \lim_{x \rightarrow a} f'(x) = \lim_{x \rightarrow a} \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

[tex] \lim_{h \rightarrow 0} \lim_{x \rightarrow a} \frac{f(x+h)-f(x)}{h} =\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=f'(a)[/tex]

But how to justify that is escaping me right now.

[tex] \lim_{x \rightarrow a} f'(x) = \lim_{x \rightarrow a} \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

[tex] \lim_{h \rightarrow 0} \lim_{x \rightarrow a} \frac{f(x+h)-f(x)}{h} =\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=f'(a)[/tex]

But how to justify that is escaping me right now.

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- #6

antibody

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It seems like you should not, because

D = lim(x->a) f'(x) {Given}

= lim(x->a) lim(y->x) (f(y)-f(x))/(y-x) {Definition of derivative}

= lim(a+h->a) lim(a+h+g->a+h) (f(a+h+g)-f(a+h))/g {Substitution}

= lim(h->0) lim(g->0) (f(a+h+g)-f(a+h))/g {Arithmetic}

= lim(h->0) lim(g->0) ((h+g)f'(a) - h f'(a)) /g {Definition of derivative}

= f'(a) lim(h->0) lim(g->0) g/g {f'(a) is just a number, move it outside}

= f'(a) lim(h->0) 1 {Easy}

= f'(a) {Easier}

What do u guys think???

- #7

DeadWolfe

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Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?

any one can give an example? thx!

Well, wouldn't it be no by the theorem that says if f is differentiable, f' has the intermediate value property.

- #8

JasonRox

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Well, wouldn't it be no by the theorem that says if f is differentiable, f' has the intermediate value property.

It does say that? Where?

- #9

Mystic998

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- #10

mathwonk

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in fact it foollows from the rolle theoirem. if f' is positive at a and negative at b, then f is not monotone in the interval [a,b], hence f takes the same value twice, hence by rolle, f has a critical point on that interval, i.e. f' = 0 in there. this is essentially the IV property for f'.

i.e. f' is always zero between two points where it changes sign. you can easily dress it up to the general statement.

- #11

ZioX

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Some of you guys are being sloppy, so I should say this for everyone browsing this thread who may get the wrong idea. f' (that is the derivative) has the intermediate value property (Darboux property) if f' is defined on an **interval**.

You can easily construct a counterexample to the original claim, just define f as straight lines pieced together so that f is continuous. It is clear that f' only takes on the values of the slopes m1, m2, m3,... etc for each line segment. Hence f' does not have an intermediate value property. This doesn't contradict f' darboux property, as f' wasn't defined in an interval (note how if we restricted f to one line segment we have f' defined and would have the intermediate value property (f'=m) which isn't very interesting).

Look to see if you can find counterexamples where f' isn't defined on an interval. This is the key to proving your problem.

You can easily construct a counterexample to the original claim, just define f as straight lines pieced together so that f is continuous. It is clear that f' only takes on the values of the slopes m1, m2, m3,... etc for each line segment. Hence f' does not have an intermediate value property. This doesn't contradict f' darboux property, as f' wasn't defined in an interval (note how if we restricted f to one line segment we have f' defined and would have the intermediate value property (f'=m) which isn't very interesting).

Look to see if you can find counterexamples where f' isn't defined on an interval. This is the key to proving your problem.

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- #12

JasonRox

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Some of you guys are being sloppy, so I should say this for everyone browsing this thread who may get the wrong idea. f' (that is the derivative) has the intermediate value property (Darboux property) if f' is defined on aninterval.

Exactly! Darboux's Theorem or Property whichever way you remember it.

The interval being (a,b) instead of [a,b].

- #13

HallsofIvy

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- #14

mathwonk

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secondly "darboux's" theorem is almost trivial as i indicated aboive:

if f'(a) <0 and f'(b)>0, we want to show f'=0 in between. but clearly f is not monotone on [a,b], hence it takes the same value twice (by IVT), and since assumed differentiable in between a and b, there is a point between where f'=0. bingo.

- #15

antibody

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so confused,,,some ppl say it does, some ppl say it does not ..

anyone can give a conclusion?

anyone can give a conclusion?

- #16

mathwonk

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in the questions posed, it only makes sense (because f is differentiable at a) to assume that f is defined and differentiable on an open interval containing a.

in that situation, if the limit of f'(x) exists as x-->a, then thnat limit must equal f'(a).

the reaqson is that if not, then there would eventually be an interval (a-e,a+e) such that on the punctured interval (excluding a) f' would be near D, but at a f'(a) would be far from D.

this would violate the IV propeerty for derivatives, since f' would not take on all values between f'(a) and the values f'(x) near D.

so the function with f'(a) * D cannot exist, because of the IV property for derivatives. It remains to oprove thaty proeprty, "Darboux's" theorem, which is a trivial corollary of rolles thm. we do this next.

- #17

mathwonk

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proof: by definition of derivative, there must be point to the right of a where f(x) > f(a), and points to the left of b where f(x) > f(b). hence f is not monotone on [a,b].

hence by the IVT for f, which is continuous, f takes the same value twice between a and b.

then at a local extremum for f on the closed interval with those two points as endpoints, there is a place where f'=0 between a and b.

now suppose f'(a) < K and f'(b) > K. then f-Kx has derivative zero between a and b, i.e. f has derivative K between a and b. so the IV proeprty of f' is proved, and the problem is completely solved.

go show your professor how easy it is, but practice it on your roommate first.

the fact that this sort of thing is unknown even to some professors, is a result of the subject being taught thoughtlessly year after year, with every book exactly like all the other books, without the authors thinking about what is happening.

- #18

JustinLevy

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so confused,,,some ppl say it does, some ppl say it does not ..

anyone can give a conclusion?

I think I caused some of the confusion by misreading your question. I gave an example where f'(a) does not equal the limit of f'(x) as x->a and that limit exists.

This however, as was pointed out, is NOT was you originally asked:

I think the OP is asking if there's an example where both the limit as f'(x) approaches aand f'(a) exist, but are not equal.

In the example I gave, f'(a) doesn't exist.

If f'(a) does exist, everyone here is in agreement that it must be equal to the limit of f'(x) as x->a. And even better, some here have constructed short proofs of this.

- #19

ZioX

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I'm just practising my LaTeX. Mathwonks proof is completely sufficient.

**The Darboux Property of Derivatives.** Let [tex]f[/tex] be differentiable on the interval I. Suppose a and b are in I and there exists an m in I such that m is strictly between [tex]f'(a)[/tex] and [tex]f'(b)[/tex], then there exists a c strictly between a and b such that [tex]f'(c)=m[/tex].

*Proof.* Assume WLOG that a is strictly less than b and [tex]f'(a)<m<f'(b)[/tex]. Define [tex]g(x)=f(x)-mx[/tex]. Then [tex]g'(a)=f'(a)-m<0[/tex] and [tex]g'(b)=f'(b)-m>0[/tex]. Since [tex]g[/tex] is continuous it follows that g must have a minimum on [a,b] (continuity on a compact set). Now, c cannot be a as [tex]g(x)\ge g(a)[/tex] so that for all x in (a,b]

Similarly c is not b. Therefore, c is in the interior of [a,b]. But then [tex]g'(c)=0[/tex] so that [tex]f'(c)=m[/tex].

This result is really interesting. It says even though the derivative might not be continuous it still has the intermediate value property.

[tex]\[\frac{g(x)-g(a)}{x-a}\ge0\][/tex]

which would imply [tex]g'(a)\ge 0[/tex].Similarly c is not b. Therefore, c is in the interior of [a,b]. But then [tex]g'(c)=0[/tex] so that [tex]f'(c)=m[/tex].

This result is really interesting. It says even though the derivative might not be continuous it still has the intermediate value property.

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- #20

mathwonk

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- #21

ZioX

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[tex]f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}=\lim_{x \to a} f'(x)=D[/tex]

using L'hopital.

- #22

mathwonk

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- #23

ankur162

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i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..

f(x)=(x+1)/(x+1) it exits Lhl=rhl at x->a just check out why to find derivative

- #24

eaboujaoudeh

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know the L'Hopital theorem?? can't we use it here?

- #25

ankur162

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know the L'Hopital theorem?? can't we use it here?

no we can't use L hospital rule here for its application limit must be having 0/0 or infinity/infinity form .

for eg: lmt x-> 1 (x-1)/2x^2-7x+5 , this can be solve easily using L hospital since it has o/o form ,

just check it on keeping limit

but u can not solve f(x)= lmt x->a (x+1)/(x+1) using L hospital as it is not a o/o form

- #26

Office_Shredder

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- #27

Robokapp

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Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?

any one can give an example? thx!

let me give it my shot:

call f'(x) = g(x) /* Why not */

lim_x->a g(x)=D

g(a) != D

So...lim_x->a g(x) = g(a) if the function is continuous...obviously.

lim_x->a g(x) = D if the function is defined at that area.

How about

{g(x) = x/x for all x !=0

g(x) = 0 for x=0}

It's a piecewise function. So we're concerned about a=0

Lim_X->0 g(x)=1 (because deltaX is never = 0 although it's always close to it therefore DeltaX / DeltaX = 1.

but Lim_x->0 g(x) != g(a) because g(0) is defined as 0.

obviously 0/0 != 0

--------------

so by plugging the stuff back (i still don't know hwy you use f'(x) ) you get in my example f'(x) = x/x or any other thing with a hole in it...so that the function has a limit but not a value there.

--------------

Simplified explanation:

A differentiable function at one point is defined as follows:

Lim_X->a f(x) = f(a)

I don't remember the formal deffinition but this is simply the exact opposite than it...Imagine a function perfectly continuous and fiferentiable at all points, then all of a sudden a hole in graph and function is defined over that hole as a different value.

- #28

jostpuur

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Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?

any one can give an example? thx!

Isn't

[tex]

f(x)=x^2 \sin\frac{1}{x}

[/tex]

a standard example of this? If you assume [tex]f(0)=0[/tex] so that the function becomes continuous.

[tex]

\lim_{x\to 0} f'(x) \neq f'(0)

[/tex]

EDIT: I just realized my mistake... :( Don't bother correcting this.

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- #29

jostpuur

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[tex]f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}=\lim_{x \to a} f'(x)=D[/tex]

using L'hopital.

I checked the assumptions in the L'hospital rule. You must assume that [tex]\lim_{x\to x_0} f'(x)/g'(x)[/tex] exists, and the the theorem says that this limit is equal to [tex]\lim_{x\to x_0} f(x)/g(x)[/tex]. (And then there was a bunch of other assumtions, but this one was the most relevant now.)

- #30

bri_guy

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I know this is old but for anyone interested.....

f(x)=(x^2)sin(1/x) if x isn't 0

f(x)=0 if x is 0.

You can show that f(x) is continuous by squeeze theorem.

recall to prove continuity you must show lim x->c on f(x)=f(c)

Now you wanted to see a derivative that exists at a point but is not continuous there.

f'(x)=2xsin(1/x)-cos(1/x) if x is not 0...notice lim x->0 on f'(x) DNE so it can't be continuous at x=0. But does the derivative exist at x=0

recall the def of a derivative...lim h->0 [f(0+h)-f(0)]/h is the derivative at 0.

lim h->0 [h^2*sin(1/h)-0]/h

lim h->0 hsin(1/h)=0 after applying squeeze theorem.

so...f'(0)=0, but lim x->0 f'(x) DNE so f(x) is differentiable at 0, but is not continuous there....in fact it is differentiable everywhere but just isn't continuous at x=0

http://www.math.tamu.edu/~tvogel/gallery/node5.html

says the same

- #31

jostpuur

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Haha. This makes me happy. I wasn't the only one making the mistake

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