Re: derivative problem correctness check

On 2006-12-01 20:51:35 -0500, "W. Dale Hall" <wdunderscorehallatpacbelldotnet@last> said:

Kira Yamato wrote:
On 2006-12-01 20:30:23 -0500, Kira Yamato <kirakun@xxxxxxxxxxxxx> said:

On 2006-12-01 20:15:49 -0500, "jraul" <jraulinth@xxxxxxxxx> said:

Can someone help me on this problem:

Assumptions:

f has finite derivative at each point of (a,b).
f '(x) ---> L < oo as x--->c, c in (a,b).
Show that f '(c) = lim_{x-->c} f '(x)

This doesn't seem right. This seem to be a counterexample:

f(x) = x^2 sin(1/x) if x <> 0,
0 if x = 0.

Nevermind. the limit of the derivative doesn't exists for this f.

What I meant here is that
lim_{x->0} f'(x)
doesn't exist.

f'(x) = -cos(1/x) + 2x sin(1/x) for x<>0.

But his problem assumes the limit of f'(x) exists at x=c.

Sorry for the many confusions.


Sorry.


Of course it does: f'(0) = 0.

The graph of f(x) lies between the two parabolas

y = x^2 and y = -x^2

intersecting each one infinitely often as x --> 0.
Because of this, any secant line joining (0,0) to
a point (x,y) on the curve must lie between the
corresponding secant lines joining (0,0) to the
two parabolas, at (x,x^2) and (x,-x^2). In particular,
the slope of the secant to (x,y) is constrained to
lie between the values -x and x. Now, let x --> 0,
and note that the secant's slope must approach 0.

Dale

--

-kira

.

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