Substituting 1 in the given quotient results indeterminate solution 0/0
Finding the limit of the rational function is determined by factorizing the numerator and denominator .
(u^4-1)/(u^3+5u^2-6u)
Let us factorize u^4-1 by applying the difference of two squares
color(blue)(a^2-b^2=(a-b)(a+b))
u^4-1=color(blue)((u^2)^2-1^2)
u^4-1=color(blue)((u^2-1))(u^2+1)
u^4-1=color(blue)((u-1)(u+1))(u^2+1)
color(red)(u^4-1=(u-1)(u+1)(u^2+1)
Factorization of u^3+5u^2-6u
u^3+5u^2-6u=u(u^2+5u-6u)
We can apply the trial and error theorem that is by finding two integers whose sum is +5 and product-6
These two integers are :-6 and +1
u^3+5u^2-6u=u(u^2+5u-6u)
u^3+5u^2-6u=u(u-(-6))(u-(+1))
color(red)(u^3+5u^2-6u=u(u+6)(u-1)
Let us compute the limit of the rational function
lim_(u->1)(u^4-1)/(u^3+5u^2-6u)
=lim_(u->1)color(red)((u-1)(u+1)(u^2+1))/color(red)(u(u+6)(u-1))
=lim_(u->1)color(red)(cancel((u-1))(u+1)(u^2+1))/color(red)(u(u+6)cancel((u-1)))
=lim_(u->1)((u+1)(u^2+1))/(u(u+6))
=((1+1)(1^2+1))/(1xx(1+6))
=(2xx2)/(1xx7)
=4/7