(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

2. Relevant equations

None

3. The attempt at a solution

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

p=0

then we have lim (x-->0+) x^{0}sin (1/x) = lim (x-->0+) sin (1/x)

Let x_{n}=1/nπ and y_{n}=1/(2n + 1/2)π

f(x_{n})= sin (1/x_{n}) = sin (nπ) n=1,2,...,n

lim (n-->∞) x_{n}= 0

lim f(x_{n}) = 0

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

p<0

lim (x-->0+) x^{p}sin (1/x)

f(x_{n}) = (1/nπ)^{p}sin (nπ)

f(y_{n}) = (1/(2n + 1/2)π)^{p}sin ((2n + 1/2)π)

lim (n-->∞) x_{n}=0

lim (n-->∞) f(x_{n})=0

We have lim (n-->∞) x_{n}=lim (n-->∞) f(x_{n})=0 so the lim (x-->0+) x^{p}sin (1/x)=0

Not too sure about this conclusion either...

p>0

Using the squeezing theorem:

-1<sin(1/x)<1

-x^{p}<x^{p}sin(1/x)<x^{p}

So if we have:

lim (x-->0+) x^{p}=0

lim (x-->0+) -x^{p}=0

Then, lim (x-->0+) x^{p}=lim (x-->0+) -x^{p}=lim (x-->0+) -x^{p}sin(1/x)=0

Thanks in advance!

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# Homework Help: Find the values of p for which the limit exists

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