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\[\lim_{x \rightarrow a}\frac{ x^{4}-a^{4} }{ x-a }\] Wolfram says it = 4a^3 I can't see how. :\
this is the derivative of the function\[f(x)=x^4\] at x=a don you see why
So just power rule it to be 4a^3, skipping any other steps?

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Other answers:

using \[\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\]
That's assuming you've already taken up derivatives...
\[=f'(x)\]
Treat \[x ^{4}-a ^{4}\] as \[(x ^{2})^{2}-(a ^{2})^{2}\] then it's algebra.
Although, if you wish to be a purist, you can factor out the numerator... \[\huge \frac{x^4-a^4}{x-a}=\frac{(x-a)(x^3+ax^2+a^2x+a^3)}{x-a}\]And the rest is dandy :D
No need for a derivative. Factor the numerator.
i was just giving for a general \[\frac{x^n-a^n}{x-a}\]
I did do 4(x-a)/(x-a) but then that just leaves 4 if x-a and x-a cancel one another.
Yeah... but \[\large \frac{4(x-a)}{x-a}\ne\frac{x^4-a^4}{x-a}\]
4 there is an exponent, not a factor ;)
Ahh yeah, I don't know why I was thinking that lol.
|dw:1364049326451:dw| From here it's simple algebra...
@Jonask And you should find that that \((x-a)\) is a factor of \(x^{n} - a^{n}\) for a wonderfully wide range of values of \(n\).
binomial theorem ie
hey that was good y did you delete
Sorry, made a typo \[\large x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+ \ ... \ +a^{n-2}x+a^{n-1})\] For as long as \[\huge n\in\mathbb{Z}^+\]
I'll remember that @terenzreignz :)
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
This is a form of the limit definition of the derivative that doesn't come up as often, \[\large \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \qquad = \qquad f'(a)\] Just something to keep in mind :) Looks like you have plenty of great assistance already though heh
yep and since \[x\to a\implies x\ne a\implies (x-a)\ne0\] you can safely cancel off the (x-a) term from numerator and denominator

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