(xn)′=h→0limh(x+h)n−(xn)=h→0limh(k=0∑n(kn)xn−khk)−xn=h→0limhxn+(k=1∑n(kn)xn−khk)−xn=h→0limhk=1∑nh(kn)xn−khk−1=h→0limhh(k=1∑n(kn)xn−khk−1)=h→0limk=1∑n(kn)xn−khk−1=h→0lim((1n)xn−1h0+(2n)xn−2h1+(3n)xn−3h2+⋯)=h→0lim(nxn−1+h((2n)xn−2+(3n)xn−3h1+⋯))=nxn−1+0((2n)xn−2+(3n)xn−301+⋯)=nxn−1
#Calculus
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I see an error in the fifth line of your reasoning to the power rule. Since you are summing the h's over and over again you can simply factor it out so once it is factored out the h no longer appears in the summand.
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Oops. Thanks for pointing out.
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No Problem. I have one more question, do you like the Joy Of Problem solving course. I seem to love it. I want to know from you if there are other problem solving courses too on this. Thanks.
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I wonder if there is any way to show this without calculus? @Gandoff Tan
@Brilliant Mathematics, @Creative Biogene is not talking about mathematics.
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