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Mathematics 25 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

factor an cancel then plug in 1

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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