Question

how to solve
lim (u^4-1)/(u^3-1)
u->1

Answer 1

Factor the top and bottom (both can factor out a u-1) then evaluate at u=1.

Answer 2

From Hopital's rule we have:
lim (u^4-1)/(u^3-1) = lim 4u^3/3u^2= lim 4u/3 =4/3

Answer 3

to civil.ad89 if we don't use hospital's rule... how can we solve it? btw.. to mtbender74 but wat is got is wrong as when u take u-1 out, then the fraction is more complicated after that.. but if i was wrong.. do you mind show me the steps?

Answer 4

factor an cancel then plug in 1

Answer 5

numerator is
\[\frac{u^4-1}{u^3-1}=\frac{(u+1)(u-1)(u^2+1)}{(u-1)(u^2+u+1)}\]

Answer 6

you get
\[\frac{(u+1)(u^2+1)}{u^2+u+1}\]

Answer 7

put
\[u=1\] get
\[\frac{4}{3}\]

Answer 8

if we don't use hospital's rule:
lim (u^4-1)/(u^3-1)= lim [(u^2-1)(u^2+1)]/(u-1)(u^2+u+1)= lim [(u-1)(u+1)(u^2+1)]/ (u-1)(u^2+u+1)= 4/3