Calculus- Continuity on intervals?

Question

The function f is defined for all x that do not equal zero, in the interval -1 f(x)= (x-sin2x)/(sinx).

A) How should f(0) be defined in order that f be continuous for all x in the interval -1
B) With f(0) defined as in part (a), use the definition of the derivative to determine whether f'(0) exists. (I can probably do this part but I need some help starting this equation.)

And for those of you that have noticed yes I have been sitting here doing calculus for about 6-7 hours :P THANKS EVERYONE that has been helping out on some of this stuff, I'm still trying to get a hold on some of these concepts.

Answer 1

OK.
In the first part, x = 0 is excluded since it would be undefined. But 0 is contained in the open interval (-1,1)

To do part A, set up the numerator and denominator in terms of the Taylor series for sine. It's probably best to split it up as x/sin(x) - sin(2x)/sin(x) since that will make it easier to 'see'.

The part that seems to cause the most grief to the most students is that the function exists everywhere *except* at x = 0 and it's well defined and well-behaved. But at x = 0 it simply 'vanishes' since 0/0 is undefined.

Think of walking down a garden path set with round 'stepping stones'. Imagine these are the points on the function. (Yeah, I know, it's a poor picture, but bear with me ☺) You suddenly come to a spot where a stone is missing. But you know where that stone would be *if* it was there. That's the entire idea behind limits. You know what it would be *if* you ever got there.

Once you've got that part done, the rest is easy (well.... sorta ☺) Just use the lim σ->0 of (f(x+σ) - f(x))/σ to get the derivative of f at the point x = 0 and you're done.

Kewl??


Doug

P.S. Tragictrust only thinks he knows what he's doing ☺

Answer 2

Part A is asking you to find a new function that is defined when x = 0 and is equal to f at all the other points. You can do this by finding the limit of f as x approaches zero. Then you can define the function piecewise.
f = (x-sin2x)/sinx for -1 f = (the limit that you found above) for x = 0

Answer 3

in A, what you are being asked to do, essentially is to define a piecewise function so that it is continuous at zero. the function f is not defined at zero, but you can simply assign f(0) a value so that the function becomes continuous.

simply looking at the graph of the function, or left and right limits at zero, you should intuit, that f(0) should be -1. thus define

g(x) = f(x) if x =/= 0, and -1 if x =0

one you have this, you should be able to figure out how to take the derivative (with standard limits).

Answer 4

The only problem I see in this interval could be when denominator = 0

sin x = 0

then x =0
at that point the numerator is also zero

lim x-> 0 (x-sin 2x)/sin x = lim x->0 (x/sin x - 2cosx sinx/sinx)
= lim x->0 x/sin x - 2cos x
= 1-2 (as lim x->0 x/sin x = 1)
= -1
so if f(0) = -1 then it is continuous

Answer 5

Do part B first.

If you differentiate the numerator and denominator separately you can find a limit as x approaches 0 that makes the function continuous at 0:
lim(1-2cos2x)/cos x) as x -->0 = -2

By definition, this is also lim((x-sin2x)/(sin x)) as x-->0,
so you may define
f(0) == -2
to make your function(s) continuous for all x.

(Remember, that in rigorously calculating limits, you never let the variable reach the limit value exactly.)

You can demonstrate this by induction by solving the function for x=0.01, 0.001, 0.0001, 0.00001, . . . .
since both x and Sin x are symmetrical about 0, the positive value proof demonstrates the negative value proof.

Answer 6

a) f(0) = lim[as x->0]((x-sin2x)/(sinx)) ...


break it into two pieces

(x/sinx) - (sin2x / sinx)

using L'Hopital's Rule":

D(x)/D(sinx) - D(sin2x) / D(sinx) =

1/cosx - 2cos2x / cosx =

(...at x=0 ...)

1/1 - 2*1/1 = 1 - 2 = -1 qed I'm pretty sure

A) f(0) = -1

............

B)
f'(0) .. ??

f(x) ( ..... as NOW defined) is a continuous function

let R = (x-sin2x)
let S = (sinx)

so f(x) = R/S

then f'(x) = (Sd(R) - Rd(S) ) / S^2

f'(0) = (Sd(R) - Rd(S) ) / 0 ... doesn't "exist"

unless (Sd(R) - Rd(S) ) = 0
at x=0 {"do the math" lololol}

that should get you there

peace

Answer 7

You need to put in zero (0) in the formula

f(0) = (0 - sin 2(0))(sin0)

You also need to find the derivative of the formula and put in zero in that formula.

Another thing to note, I believe you also need to put in one and negative one too.

It has been a couple of years since I completed my Calculus class in college, so I think I am right.