Easy limit

... which I cannot do :)
lim (x->inf) x^(sin(x))

simple, there is no limit, because x^sin(x) does not converge as x gets infinitely big

Interesting function - took me a while to imagine what that looks like.

BugZ is right in that it does not converge. Its values, I believe, range from ~0 to infinity. When sin(x) = -1, then the limit would be 0, but when sin(x) = 1, the limit would be infinity. (Or, as BugZ might call it, "last seen heading north" or LSHN :D) However, sin(x) continues to oscillate at a steady pace, thus x^sin(x) doesn't converge.

Destined

Well folks, it looks like this:

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Originally posted by Destined Soul
Interesting function - took me a while to imagine what that looks like.

BugZ is right in that it does not converge. Its values, I believe, range from ~0 to infinity. When sin(x) = -1, then the limit would be 0, but when sin(x) = 1, the limit would be infinity. (Or, as BugZ might call it, "last seen heading north" or LSHN :D) However, sin(x) continues to oscillate at a steady pace, thus x^sin(x) doesn't converge.

Destined

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lol, well i would call it no limit, not LSHN, because x is approaching infinity. however, if x is appoaching c (some constant) and the limit is infinity, only then I'd call it LSHN.

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Originally posted by A$$Bandit
Well folks, it looks like this:

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when you sin(x), are you treating x as radians, or as degrees,
since x^sin(x) would appear different under those circumstances.

example:

(360)^sin(360) would be 1 if using degrees,
and would be 282.669 if using radians.

True, you could use degrees, but isn't it usually implied that radians are used? Either way, the plot will still look the same, but just with a different scale.

Destined

Sorry guys, slight mistake in the limit :)
lim (x->0+) not inf
Makes a slight difference I think ;)

Ah.. much better.

sin(x) = x - x³/3! + x⁵/5! - ...

So if |x| << 1, then sin(x) ~ x, and lim(x-->0+) of x is 0.

Anything to the power of 0, (I think anything, not sure exactly) is =1. Thus 0⁰ = 1.

Of course, someone else might better explain the x⁰ = 0 for what values of x. :p

Destined

I believe I used radians, but as someone said it only affects the horizontal scale really...

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Originally posted by Destined Soul
True, you could use degrees, but isn't it usually implied that radians are used?

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I'm not sure, but I think for casual use, it is dependent on the standard units of measure in your local culture, and is not written in stone.

However, it more likely is so in the scientific and mathematical professions.

But, I believe I made a valid observation, either way.

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Originally posted by Destined Soul

Either way, the plot will still look the same, but just with a different scale.

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So you are saying:

X^{Sin(Xin radians)} = ScaleFactor*X^{Sin(Xin degrees)}

Or, in other words, ScaleFactor is a constant for all X in the following:
ScaleFactor = (X^{Sin(X in radians)})/(X^{Sin(X in degrees)})

:confused:

Sorry, as usual, I'm doing the math in my head instead of on paper.

For some value x in degrees, and its equivalent y in radians, both are raised to the same power, since sin can be operated in both degrees and radians. (Well, sin(x) does oscillate between -1 and 1 and a set frequency = 2*Pi radians or 360 degrees.)

Thus for some x and y, you have x^a and y^a for some value. These are related in that if you took the power of 1/a of each of these, then they are scalable.

I just used the wrong words. I think the word is very similar. :D

Hopefully this sets some of what I said. In sum: I hate english. :p It's too confusing.

Destined

the graph f(x)=a^(sinx) compared to g(t)=a^(sin(t)) where t is expressed in degrees experiences a horizontal compression of pi/180. when its x^(sinx) its different but i am not sure how different.

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Originally posted by Destined Soul
Ah.. much better.

sin(x) = x - x³/3! + x⁵/5! - ...

So if |x| << 1, then sin(x) ~ x, and lim(x-->0+) of x is 0.

Anything to the power of 0, (I think anything, not sure exactly) is =1. Thus 0⁰ = 1.

Of course, someone else might better explain the x⁰ = 0 for what values of x. :p

Destined

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No, I don't think 0[sup]0 is 1. Can't remember the exact reason. I do know that you're supposed to use L'Hospital's rule for this problem.

Btw how did you get sin(x) = x - x[sup]3/3! + x[sup]5/5! - ...

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Originally posted by marnitzg
Btw how did you get sin(x) = x - x3/3! + x5/5! - ...

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this is the sin(x) expansion of the Taylor series, although i don't see how it helps. we've probably established that when x=0, sin(x)=0. also, 0⁰ is indeterminate, as you will see an error if you punch in to any calculator. however, keep in mind that we are not evaluating 0⁰, but we are evaluating a limit, so technically saying that the limit (x->0+)x^sin(x)=1 is ok i guess, and it can be varified by plugging into a very small value (of course, x is understood to be radians)

Ok, worked something out. Just check my working pls.
x^sinx = e^sinx*lnx
As e is continous we can interchange the limit right?
=sinx*lnx
=lnx/cscx
By L'Hospital's theorem
=1/x/(-cscx*cotx)
=-1/(x*cscx*cotx)
=-sin² x/(x*cosx)
Differentiating again
=(2sinx*cosx)/(cosx*sinx)
=2
therefore = e²

Or did I do something wrong?

i lost you in the first two lines.

He he. Maybe this will help:
ln(x) = log_e(x)
e^ln(x) = x
ln(x)^y = y * ln(x)

Therefore
x^sinx = e^{ln(x^(sinx))} = e^{sinx * lnx}

nope next line. sorry should've been more specific. i have no clue what u meant by

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As e is continous we can interchange the limit right?

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are you saying that you are trying to find the limit (x->0) of e^sinx*lnx? if so what happened to the e?

e^x is continous for all values of x
Therefore
lim(x->0+)e^x = e^lim(x->0+)x

If you can't interchange the limit with the exponent then there
is surely no way of calculating this limit?

I don't get it either.

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Originally posted by bugzpodder
and thats wrong also. should be:

cscx*cotx/x

or cosx/(xsin^2(x))

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Err, sorry.
Notation should read: (1/x)/(-cscx*cotx)
=(1/x)*(1/(-cscx*cotx))
=-(1/(x*cscx*cotx))

Just a slight misread of the notation. This is what it should
look like, but the answer still remains the same

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Originally posted by bugzpodder
ah ok, yes i see what u mean. should've write that. but the e is still there. let me continue ur work:

=lim(x->0) e^(cosx/(xsin^2(x)))

you'll never get the sin^2(x) out of the bottom btw. don't try L'Hopital's theorem again

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Yeah, the e is still there. Its just hard to display it on an html page
so I left it out. You'll notice that I do bring it back in the end

You'll notice from my previous post that you're interpretation is incorrect. Therefore my division still holds.
BTW what do you mean by "you'll never get the sin^2(x) out the bottom?

yes of course.

=lim(x->0) -sin^2 x/(x*cosx)
should be
-2sinxcosx/(cosx-sinx)

is it just me today?? am i wrong again? btw forget about 1/x and sin^2(x) i already deleted my post it was a misunderstanding

Now you've lost me. How did
-sin² become -2sinxcosx
and x * cos x
= cosx - sinx
:confused: * 1000

no the derivitive of -sin^2(x)=-2sinxcosx
and the derivitive of xcosx=-sinx+cosx

am i right? i am kinda confused too

I think you got confused with
sin2x = 2sinxcosx

wrt xcosx
By the chain rule d/dx f(x)g(x)
= f'(x)*g(x) + g'(x)f(x)
therefore f(x) = x and g(x) = cosx
therefore d/dx = 1*cosx + -sinx * x
Right?

Which disproves my earlier workings :)

Ok, the revised version

x^sinx = e^sinx*lnx
As e is continous we can interchange the limit right?
=sinx*lnx
=lnx/cscx
By L'Hospital's theorem
=(1/x)/(-cscx*cotx)
=-1/(x*cscx*cotx)
=-sin² x/(x*cosx)
Differentiating again
=-(2sinx*cosx)/(cosx - xsinx)
Now taking the limit
=-(2*0*1) / (1 - 0 * 0)
=0/1 = 0

therefore = e⁰ = 1

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Originally posted by marnitzg
I think you got confused with
sin2x = 2sinxcosx

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no

Quote:

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wrt xcosx
By the chain rule d/dx f(x)g(x)
= f'(x)*g(x) + g'(x)f(x)
therefore f(x) = x and g(x) = cosx
therefore d/dx = 1*cosx + -sinx * x
Right?

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called the product rule but yes, your right (hehe forgot about x)

and double check ur derivitive of -sin^2(x)

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Originally posted by marnitzg
Ok, the revised version

x^sinx = e^sinx*lnx
As e is continous we can interchange the limit right?
=sinx*lnx
=lnx/cscx
By L'Hospital's theorem
=(1/x)/(-cscx*cotx)
=-1/(x*cscx*cotx)
=-sin² x/(x*cosx)
Differentiating again
=(2sinx*cosx)/(cosx - xsinx)
Now taking the limit
=(2*0*1) / (1 - 0 * 0)
=0/1 = 0

therefore = e⁰ = 1

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derivitive of -sin²x
is -2sinxcosx, although it won't change your answer. after you fix it it will be better

Sorry, was thinking of the chain rule for sin²x
as sin²x = (sin x)²
= 2* d/dx (sin x) = 2 * -cos x

I'm positive about this now cause I just looked it up in my calc textbook :)

Quote:

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Originally posted by marnitzg
Sorry, was thinking of the chain rule for sin²x
as sin²x = (sin x)²
= 2* d/dx (sin x) = 2 * -cos x

I'm positive about this now cause I just looked it up in my calc textbook :)

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nope still don't think so. derivitive of -sin^2(x)=-2cosxsinx

Sorry, you are right. It was very late last night! I had that written down on paper but for some reason it looked wrong :). Hell, I even looked the wrong sum up in the textbook!

Btw, agreed then that the limit is 1?

yes :D

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Originally posted by marnitzg
Btw, agreed then that the limit is 1?

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wasnt that just obvious?:D
anything to the power 0 is 1 (or not:p)

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Originally posted by MrPolite
wasnt that just obvious?:D
anything to the power 0 is 1 (or not:p)

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No, think 0⁰ is undefined as in can represent any number, just like 0/0. Also think about infinity to the power of zero.

0⁰ is indeterminate. in case of a limit, it means that you have to do clever algebraic manipulation to the expression to get a valid answer -- although it does mean most likely that the limit exists (normally i consider L'Hopital's method a sledge hammer or machine gun that beats up the question, but in this case i don't see another way out of this without L'Hopital's method). undefined value however is more like infinity (positive/negative)

LOL. It works does it not? Which theorems do you prefer to use? We're only going to do Taylor's theorem in our next class

ahh any function can be written as the a taylor series. whats the Taylor's theorem good for?

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Originally posted by bugzpodder
ahh any function can be written as the a taylor series. whats the Taylor's theorem good for?

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what's taylor's theorem?:confused:

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Originally posted by bugzpodder
ahh any function can be written as the a taylor series.

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Originally posted by MrPolite
what's taylor's theorem?:confused:

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Taylor's Theorem states:
Suppose that f has n+1 continous derivatives on an open interval
I containing 0. Then, for each x E I,
f(x) = f(0) + f'(0)x + (f''(0)x²)/2! + ... + (f⁽ⁿ⁾(0)xⁿ)/n! + R_n+1(x),
where the remainder R_n+1 is given by the formula
R_n+1(x) = 1/n! * integral(a=0 b=x)f⁽ⁿ⁺¹⁾(t)(x-t)ⁿdt

what is it used for?

Hi Bugz. AFAIK:

a) I think in calculators/computers for calculating functions like e^x
b) Integrating functions that cannot normally be integrated
c) Generally making approximations