Is the answer for #lim_(x->2)(x+1)/(x^2-x-2# 0?

#lim_(x->2)(x+1)/(x^2-x-2#

I got 0... is that right?

1 Answer
Shray Patel
Mar 21, 2018

No. The limit doesn't exist or is undefined. It approaches negative infinity from the right and positive infinity from the left.

Explanation:

Let's try evaluating this limit algebraically first:
#lim_(x->2)(x+1)/(x^2-x-2)#
#=lim_(x->2)(x+1)/((x+1)(x-2))#
#=lim_(x->2)1/(x-2)#
#= 1/0#

Hmm...that doesn't seem to work. Let's evaluate the value of the function at values extremely close to 2, like 1.999 and 2.001
Desmos
Looks like the function approaches different values from the left and right side of 2. This means it's extremely likely the limit doesn't exist. Looking at a graph confirms this: the function approaches a value of negative infinity from the left of #x=2# and positive infinity from the right side of #x=2#:

Desmos

It seems to me that your mistake came from evaluating the function incorrectly by doing this:
#lim_(x->2)(x+1)/(x^2-x-2)#
#=lim_(x->2)(x+1)/((x+1)(x-2))#
This is where I think you might have messed up:
#=lim_(x->2)(x-2)#
#= 0#

Again, the correct answer should be:
#lim_(x->2)(x+1)/(x^2-x-2)#
#=lim_(x->2)(x+1)/((x+1)(x-2))#
#=lim_(x->2)1/(x-2)#
#= 1/0#
Therefore, the limit does not exist or is undefined at #x=2#.

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