How do you prove the statement lim as x approaches 2 for #((x^2+x-6)/(x-2))=5# using the epsilon and delta definition?

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Answer 1 · 2015-10-16T16:33:53

See the explanation below.

Preliminary Analysis

First look at #abs(f(x)-L)#. I want to make that less than #epsilon#.

I am looking for #abs(x-2)# because, through my control of #delta#, I control #abs(x-2)#.

#abs(f(x)-L) = abs((x^2+x-6)/(x-2)-5)#

# = abs(((x+3)(x-2))/(x-2)-5)#

# = abs((x+3)-5)# for all #x != 2#

# = abs(x-2)# for all #x != 2#

We want to make #abs(f(x)-L)# less than a given #epsilon#.

If we make #abs(x-2) < epsilon# BUT #x != 2#, then we will have what we want. (We cannot allow #x=2# because #2# is not in the domain of #f#.

So now, we are ready to write the proof.

Proof

Given #epsilon > 0 # choose #delta = epsilon#. (Clearly, then, #delta > 0# as required.

Now if #x# is chosen so that

#0 < abs(x-2) < delta#, then we will have

#abs((x^2+x-6)/(x-2)-5) = abs(((x+3)(x-2))/(x-2)-5)#

# = abs((x+3)-5) = abs(x-2) < delta = epsilon#.

That is:
If #0 < abs(x-2) < delta#, then #abs((x^2+x-6)/(x-2)-5) < epsilon#.

By the definition of limit,

#lim_(xrarr2)(x^2+x-6)/(x-2) = 5#