How do you find the Limit of #lim_(yto3)(y^3-13y+12 )/( y^3-14y+15)#?

Question

4654 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-10T17:54:50

Because the expression evaluated at the limit results in the indeterminate form #0/0#, the use of L'Hôpital's rule is warranted.

Given: #lim_(yto3)(y^3-13y+12 )/( y^3-14y+15)#

#lim_(yto3)((d(y^3-13y+12 ))/dx)/((d( y^3-14y+15))/dx) = #

#lim_(yto3) (3y^2-13)/(3y^2-14) = 14/13#

So goes the original limit:

#lim_(yto3)(y^3-13y+12 )/( y^3-14y+15) = 14/13#

Answer 2 · 2017-07-10T18:08:25

# 14/13.#

The Nr. =#y^3-13y+12.# and, the Dr. =#y^3-14y+15.#

Note that, in the polynomial of the Nr., the sum of the co-effs. is #0.#

# :. (y-1)# must be a Factor.

#:." The Nr." =(y-1)(y^2+y-12)=(y-1)(y+4)(y-3).#

The poly. in the Dr. vanishes for #y=3. :. (y-3)# is a Factor.

#" The Dr. "=(y-3)(y^2+3y-5).#

# :." The Lim. "=lim_(y to 3) {cancel((y-3))(y-1)(y+4)}/{cancel((y-3))(y^2+3y-5)},#

#=lim_(y to 3) {(y-1)(y+4)}/(y^2+3y-5),#

#={(3-1)(3+4)}/(3^2+3*3-5),#

# rArr" The Lim. "=14/13.#