Question

If m>1 then x^m sin(1/x^n) will be differentiable at 0

However, why?

Q1. if we take limit x to 0 sin(1/x^n), we get sin(1/0) it doesn't make sense even if m>1

Q2. why when M=1, it will not differentiable at 0?

Q3 Please tell me that limit x to infinity sin(1/x) DNE becasue limit will be -1 and 1???? is that the reason?

Answer #1

Show all the steps and explain. Don't skip steps and please
clear hand written
f(x)=x^m sin(1/x^n) if
x is not equal 0 and f(x)=0 if x =0
(a) prove that when
m>1+n, then the derivative of f is continuous at 0
limit x to 0 x^n
sin(1/x^n) does not exist? but why??? please explain it should be
0*sin(1/x^n)

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true we needed to show that the sum of the real parts equals 0 and
the sum of the imaginary parts is equal to 0.
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