How do you evaluate the limit #(2x+4)/(x^2-3x-10)# as x approaches #2#?

2 Answers
Jim G.
Oct 3, 2016

#-2/3#

Explanation:

Since this function is not indeterminate at x = 2, we can evaluate it by direct substitution.

#lim_(xto2)(2x+4)/(x^2-3x-10)=(2(2)+4)/(2^2-3(2)-10)=8/(-12)=-2/3#

Jim H
Oct 3, 2016

#lim_(xrarr2)(2x+4)/(x^2-3x-10) = -2/3# and #lim_(xrarr-2)(2x+4)/(x^2-3x-10) = -2/7#

Explanation:

As #x# approaches #2#, the numerator goes to #8# and the denominator goes to #4-6-10 = -12#, so the limit is #8/-12 = -2/3#

As #x# approaches #-2#, both the numerator and denominator go to #0#. Since boith are polynomials and #02# is a zero, we can be sure that #x-(-2)# = x+2# is a foctor of both the numerator and the denominator.

We'll factor and reduce.

#lim_(xrarr-2)(2x+4)/(x^2-3x-10)=lim_(xrarr-2)(2(x+2))/(x+2)(x-5))#

# = lim_(xrarr-2)2/(x-5)#

# = 2/((-2)-5) = -2/7#

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