How do you find the limit of #(3x^2-x-10)/(x^2+5x-14)# as x approaches 2?

Question

11490 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-30T15:25:52

#11/9#

Observe that
#3x^2-x-10=(3x+5)(x-2)#
#x^2+5x-14=(x+7)(x-2)#
So we get

#lim_(x to 2)(3x^2-x-10)/(x^2+5x-1)=lim_(x to 2)(3x+5)/(x+7)=11/9#

Answer 2 · 2018-05-30T15:31:32

#lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = 11/9#

Method 1): factorize the polynomials

As both numerator and denominator vanish for #x=2# they can be divided by #(x-2)#:

#3x^2-x-10 = (3x+5)(x-2)#

#x^2+5x-14 = (x+7)(x-2)#

Then:

#lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) ( (3x+5)(x-2))/((x+7)(x-2))#

#lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) (3x+5)/(x+7)#

#lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = 11/9#

Method 2): L'Hospital's rule:

As both numerator and denominator vanish for #x=2#:

#lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) (d/dx (3x^2-x-10))/(d/dx (x^2+5x-14 ) )#

#lim_(x->2) (3x^2-x-10)/(x^2+5x-14 ) = lim_(x->2) (6x-1)/(2x+5) = 11/9#