What's the limit of #(x^n-a^n)/(x-a)# as #x# approaches #a# , using the derivate number which is #f'(a)# ?

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Answer 1 · 2016-12-03T20:42:36

Please see below.

I'm not sure I understand your question. I don't know what you mean by "using the derivate number which is #f'(a)#"

If I understand it, I think I need to point out that

#lim_(xrarra)(x^n-a^n)/(x-a) = f'(a)# for #f(x) = x^n#.

That is:

For #f(x) = x^n#,

#f'(a) = lim_(xrarra)(f(x) - f(a))/(x-a) = (x^n-a^n)/(x-a)#

We also know, by the power rule for derivatives,

That for #f(x) = x^n#, we haved #f'(x) = nx^(n-1)# and #f'(a) = na^(n-1)#.

We can conclude that

#lim_(xrarra)(x^n-a^n)/(x-a) = na^(n-1)#