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Astrobuoy Group TitleBest 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. :\
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.2
this is the derivative of the function\[f(x)=x^4\] at x=a don you see why
 one year ago

Astrobuoy Group TitleBest ResponseYou've already chosen the best response.0
So just power rule it to be 4a^3, skipping any other steps?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.2
using \[\lim_{h \to 0}\frac{f(x+h)f(x)}{h}\]
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
That's assuming you've already taken up derivatives...
 one year ago

linshan789 Group TitleBest ResponseYou've already chosen the best response.0
Treat \[x ^{4}a ^{4}\] as \[(x ^{2})^{2}(a ^{2})^{2}\] then it's algebra.
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
Although, 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
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
No need for a derivative. Factor the numerator.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.2
i was just giving for a general \[\frac{x^na^n}{xa}\]
 one year ago

Astrobuoy Group TitleBest ResponseYou've already chosen the best response.0
I did do 4(xa)/(xa) but then that just leaves 4 if xa and xa cancel one another.
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.1
4 there is an exponent, not a factor ;)
 one year ago

Astrobuoy Group TitleBest ResponseYou've already chosen the best response.0
Ahh yeah, I don't know why I was thinking that lol.
 one year ago

linshan789 Group TitleBest ResponseYou've already chosen the best response.0
dw:1364049326451:dw From here it's simple algebra...
 one year ago

tkhunny Group TitleBest 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 \(n\).
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.2
binomial theorem ie
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.2
hey that was good y did you delete
 one year ago

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

Astrobuoy Group TitleBest ResponseYou've already chosen the best response.0
I'll remember that @terenzreignz :)
 one year ago

S.RaviTeja Group TitleBest ResponseYou've already chosen the best response.0
use 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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
This 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
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.0
yep and since \[x\to a\implies x\ne a\implies (xa)\ne0\] you can safely cancel off the (xa) term from numerator and denominator
 one year ago
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