How do you find the limit of #f(x) = (x^2 + x - 6) / (x + 3) # as x approaches 2?

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Answer 1 · 2018-04-23T12:49:28

#0#

We seek to find: #lim_(x->2)f(x)=(x^2+x-6)/(x+3)#.

Plugging in #x=2# immediately yields:

#f(2)=(2^2+2-6)/(2+3)#

#=(4+2-6)/5#

#=(6-6)/5#

#=0/5#

#=0#

So, the limit is #0# at #x=2#.

Answer 2 · 2018-04-23T12:56:37

#0#

#lim_(xto2)(x^2+x-6)/(x+3)#

#=lim_(xto2)(cancel((x+3))(x-2))/cancel((x+3))#

#=lim_(xto2)(x-2)=0#