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#
