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

Question

Elijah Abram · Accepted Answer

#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#

Ezra Adams · Answer

#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#

Samantha Adkison · Answer

undefined To find the limit of (3x^2-x-10)/(x^2+5x-14) as x approaches 2, we can substitute 2 into the expression and simplify. By substituting 2 for x, we get (3(2)^2-2-10)/(2^2+5(2)-14). Simplifying further, we have (12-2-10)/(4+10-14), which becomes 0/0. This is an indeterminate form. To evaluate the limit, we can factorize the numerator and denominator. Factoring the numerator gives (3x+2)(x-5), and factoring the denominator gives (x+7)(x-2). Canceling out the common factor of (x-2), we are left with (3x+2)/(x+7). Substituting 2 into this expression gives (3(2)+2)/(2+7), which simplifies to 8/9. Therefore, the limit of (3x^2-x-10)/(x^2+5x-14) as x approaches 2 is 8/9.