Find the values of p for which the limit exists

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In summary, the conversation discusses finding the values of p for which the limit exists in the equation lim (x-->0+) xp sin (1/x). The attempt at a solution breaks down the problem into three cases: p=0, p<0, and p>0. After analyzing each case, it is determined that the limit does not exist for p=0 due to the oscillating behavior of the sequence f(xn). This is confirmed by breaking the given sequence into two sub-sequences and showing that they approach different limits.
  • #1
frenchkiki
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Homework Statement



Find the values of p for which the limit exists: lim (x-->0+) xp sin (1/x)

Homework Equations



None

The Attempt at a Solution



Alright so I've broken it down to 3 cases:

p=0

then we have lim (x-->0+) x0sin (1/x) = lim (x-->0+) sin (1/x)
Let xn=1/nπ and yn=1/(2n + 1/2)π
f(xn)= sin (1/xn) = sin (nπ) n=1,2,...,n
lim (n-->∞) xn = 0
lim f(xn) = 0

I'm not too sure how to deduce that the limit doesn't exist here.

p<0

lim (x-->0+) xp sin (1/x)
f(xn) = (1/nπ)p sin (nπ)
f(yn) = (1/(2n + 1/2)π)p sin ((2n + 1/2)π)

lim (n-->∞) xn=0
lim (n-->∞) f(xn)=0

We have lim (n-->∞) xn=lim (n-->∞) f(xn)=0 so the lim (x-->0+) xp sin (1/x)=0

Not too sure about this conclusion either...

p>0

Using the squeezing theorem:

-1<sin(1/x)<1
-xp<xpsin(1/x)<xp

So if we have:
lim (x-->0+) xp=0
lim (x-->0+) -xp=0

Then, lim (x-->0+) xp=lim (x-->0+) -xp=lim (x-->0+) -xpsin(1/x)=0

Thanks in advance!
 
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  • #2
frenchkiki said:
...

p=0

then we have lim (x-->0+) x0sin (1/x) = lim (x-->0+) sin (1/x)
Let xn=1/nπ and yn=1/(2n + 1/2)π
f(xn)= sin (1/xn) = sin (nπ) n=1,2,...,n
lim (n-->∞) xn = 0
lim f(xn) = 0
Instead of the sequence used here, pick one that has odd number multiples of 1/(π/2), i.e., 1/(π/2), 3/(π/2), 5/(π/2), ...

I'm not too sure how to deduce that the limit doesn't exist here.
...
 
  • #3
OK then I get lim (n-->inf) x_n=+inf and
lim (n-->inf) f(x_n) =0. Is this enough to show that lim (x-->0+) sin (1/x) does not exist?
 
  • #4
frenchkiki said:
OK then I get lim (n-->inf) x_n=+inf and
lim (n-->inf) f(x_n) =0. Is this enough to show that lim (x-->0+) sin (1/x) does not exist?

It's the sequence f(xn) that's of interest here.

So, if [itex]\displaystyle x_n=\frac{1}{(2n-1)\frac{\pi}{2}}\,,[/itex] what is f(xn) ?
 
Last edited:
  • #5
f(x_n)=sin (pi n I pi/2), then I get lim (n-->inf) f(x_n)=+inf and lim (n-->inf) x_n =0
 
  • #6
SammyS said:
It's the sequence f(xn) that's of interest here.

So, if [itex]\displaystyle x_n=\frac{1}{(2n-1)\frac{\pi}{2}}\,,[/itex] what is f(xn) ?
First look at 1/xn = (2n - 1)(π/2) = π/2, 3π/2, 5π/2, ... , for n = 1, 2, 3, ...

Of course (xn)0 = 1 .

So f(xx) = sin(π/2), sin(3π/2), sin(5π/2), ... = ? ? ?
 
  • #7
SammyS said:
First look at 1/xn = (2n - 1)(π/2) = π/2, 3π/2, 5π/2, ... , for n = 1, 2, 3, ...

Of course (xn)0 = 1 .

So f(xx) = sin(π/2), sin(3π/2), sin(5π/2), ... = ? ? ?

so f(xn) is not defined because it oscillates between 1 and -1?
 
  • #8
The sequence is certainly defined. It is the limit of the sequence that is not defined.
 
  • #9
What is unclear to me is how to use the sequence to prove that the limit doesn't exist.
 
  • #10
Break the given sequence up into two sub-sequences; one with the odd terms, the other with the even terms.
 
  • #11
A brief visit at my professor's office clarified the problem. You pick two sequences xn and yn tending to 0. In the case p=0 case f(xn)=0 f(yn)=1 with xn=1/nπ and yn=1/(2n + 1/2)π.

then lim (n→∞) xn = 0 ≠ 1 = lim (n→∞) yn.

Therefore lim (x→0+) xp sin (1/x) does not exist for p=0.

Thank you both for your help.
 
  • #12
Yes, that was exactly what SammyS said.
 

FAQ: Find the values of p for which the limit exists

What is a limit in calculus?

A limit in calculus is the value that a function approaches as the input variable gets closer and closer to a specific value. It is typically denoted by the notation lim f(x), where f(x) is the function and the limit is evaluated as x approaches a specific value.

How do you determine if a limit exists?

In order for a limit to exist, the value of the function as x approaches the specific value must be equal to the limit from both the left and right sides. This means that the function must approach the same value from both directions. If this condition is met, the limit is said to exist.

What is a one-sided limit?

A one-sided limit is a limit that is evaluated from only one direction, either the left or the right. It is denoted by the notation lim f(x) and is used when a function has a different value or is undefined from one direction.

How do you find the limits of a polynomial function?

To find the limit of a polynomial function, you can simply plug in the specific value of x and evaluate the function. If the resulting value is a real number, then the limit exists. If the resulting value is undefined or approaches infinity, the limit does not exist.

Can the limit of a function exist at a point where the function is not defined?

Yes, the limit of a function can exist at a point where the function is not defined. This is because the limit is evaluated based on the behavior of the function as it approaches the specific value, not the actual value at that point. However, the function must approach the same value from both the left and right sides for the limit to exist.

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