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anonymous
 3 years ago
Limit help needed.
anonymous
 3 years ago
Limit help needed.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow a}\frac{ x^{4}a^{4} }{ xa }\] Wolfram says it = 4a^3 I can't see how. :\

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is the derivative of the function\[f(x)=x^4\] at x=a don you see why

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So just power rule it to be 4a^3, skipping any other steps?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0using \[\lim_{h \to 0}\frac{f(x+h)f(x)}{h}\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1That's assuming you've already taken up derivatives...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Treat \[x ^{4}a ^{4}\] as \[(x ^{2})^{2}(a ^{2})^{2}\] then it's algebra.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Although, if you wish to be a purist, you can factor out the numerator... \[\huge \frac{x^4a^4}{xa}=\frac{(xa)(x^3+ax^2+a^2x+a^3)}{xa}\]And the rest is dandy :D

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0No need for a derivative. Factor the numerator.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i was just giving for a general \[\frac{x^na^n}{xa}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did do 4(xa)/(xa) but then that just leaves 4 if xa and xa cancel one another.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Yeah... but \[\large \frac{4(xa)}{xa}\ne\frac{x^4a^4}{xa}\]

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.14 there is an exponent, not a factor ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ahh yeah, I don't know why I was thinking that lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364049326451:dw From here it's simple algebra...

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0@Jonask And you should find that that \((xa)\) is a factor of \(x^{n}  a^{n}\) for a wonderfully wide range of values of \(n\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey that was good y did you delete

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry, made a typo \[\large x^na^n=(xa)(x^{n1}+ax^{n2}+a^2x^{n3}+ \ ... \ +a^{n2}x+a^{n1})\] For as long as \[\huge n\in\mathbb{Z}^+\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll remember that @terenzreignz :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0use L'Hospitals rule differentiate nr. and dr. and then apply the limit becoz acc. to the rule if we apply the limit we get 0/0 so after applying the rule we get 4a^3

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.0This is a form of the limit definition of the derivative that doesn't come up as often, \[\large \lim_{x \rightarrow a}\frac{f(x)f(a)}{xa} \qquad = \qquad f'(a)\] Just something to keep in mind :) Looks like you have plenty of great assistance already though heh

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep and since \[x\to a\implies x\ne a\implies (xa)\ne0\] you can safely cancel off the (xa) term from numerator and denominator
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