## anonymous 3 years ago Limit help needed.

1. anonymous

$\lim_{x \rightarrow a}\frac{ x^{4}-a^{4} }{ x-a }$ Wolfram says it = 4a^3 I can't see how. :\

2. anonymous

this is the derivative of the function$f(x)=x^4$ at x=a don you see why

3. anonymous

So just power rule it to be 4a^3, skipping any other steps?

4. anonymous

using $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

5. terenzreignz

That's assuming you've already taken up derivatives...

6. anonymous

$=f'(x)$

7. anonymous

Treat $x ^{4}-a ^{4}$ as $(x ^{2})^{2}-(a ^{2})^{2}$ then it's algebra.

8. terenzreignz

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

9. tkhunny

No need for a derivative. Factor the numerator.

10. anonymous

i was just giving for a general $\frac{x^n-a^n}{x-a}$

11. anonymous

I did do 4(x-a)/(x-a) but then that just leaves 4 if x-a and x-a cancel one another.

12. terenzreignz

Yeah... but $\large \frac{4(x-a)}{x-a}\ne\frac{x^4-a^4}{x-a}$

13. terenzreignz

4 there is an exponent, not a factor ;)

14. anonymous

Ahh yeah, I don't know why I was thinking that lol.

15. anonymous

|dw:1364049326451:dw| From here it's simple algebra...

16. tkhunny

@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 $$n$$.

17. anonymous

binomial theorem ie

18. anonymous

hey that was good y did you delete

19. terenzreignz

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 $\huge n\in\mathbb{Z}^+$

20. anonymous

I'll remember that @terenzreignz :)

21. anonymous

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

22. zepdrix

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

23. anonymous

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