Question

Use algebra para calcular o limite
Lim 1/(1+h)-1 sobre h , isso quando H tende a 0

Answer 1

Sejam a= (h+1)^(1/3) e b=1 -->

E= [(h+1)^1/3 -1] = (a - b) = (a-b)·(a²+ab+b²)/(a²+ab+b²) =
(a³-b³)/(a²+ab+b²)= [(h+1)^(3/3) - 1³] / [ (h+1)^(2/3) + (h+1)^(1/3) + 1] =
[(h+1)- 1] / [ (h+1)^(2/3) + (h+1)^(1/3) + 1] =
h / [ (h+1)^(2/3) + (h+1)^(1/3) + 1]

lim_{h-->0} [(h+1)^1/3 -1]/h = lim_{h-->0} E/h =
lim_{h-->0} 1/[ (h+1)^(2/3) + (h+1)^(1/3) + 1] =
1/ [ (0+1)^(2/3) + (0+1)^(1/3) + 1] =
1/ [1+1+1] = 1/3