calculus
posted by Anonymous on .
evaluate lim x aproches 1
x^2+6x+5/x^23x4

lim (x^2 + 6x + 5)/(x^2  3x  4) as x> 1
Not that we can factor both numerator & denominator,
lim (x+5)(x+1) / (x4)(x+1)
We can thus cancel the x+1, leaving
lim (x + 5)/(x  4) as x> 1
Substituting x = 1,
= (1 + 5) / (1  4)
= 4 / 5
= 4/5
Hope this helps~ :3