Help with calculus problem

what is the limit of [ (1+h)^4 -1 ]/ h as h approaches 0?
Can you explain the steps also? thanks

First expand (1+h)^4 to get
(1+h)^4=(1+h)^2*(1+h)^2)=(h^2+2h+1)*(h…
=h^4+2h^3+h^2+2h^3+4h^2+2h+h^2+2h+1
=h^4+4h^3+6h^2+4h+1.
Subtract 1:
(1+h)^4-1=h^4+4h^3+6h^2+4h.
Divide by h:
[(1+h)^4-1]/h=h^3+4h^2+6h+4->4 as h->0.

Hope that helps.

Answer: 4

Method: Since lim[(1 + h)^4 - 1] = 0 = limh as h approaches 0, you can use L'Hopital's rule. L'Hopital's rule states that if both the numerator and the denominator approach 0 (or +/- infinity), you can replace the numerator and denominator with their respective derivatives and the limit of the resulting fraction is the same.

So, differentiate the numerator and denominator (separately):
[(1 + h)^4 - 1]' = 4(1 + h)^3
h' = 1

So, now just solve lim[4(1 + h)^3] as h approaches 0. Here you can just plug in 0 for h to obtain:
4(1 + 0)^3 = 4

I would use L'hospital's rule, but that depends on the use of derivatives for what looks like a difference quotient, so the argument would become circular.
Expand (1+h)^4, then subtract the one, cancel off the h's, then take the limit. Mostly a big algebra mess, but nothing too confusing.
Good Luck.