Use the precise definition of a limit to prove

Question

lim (x²+x-6)/(x-2) = 5

x-> 2

Answer 1

Well the numerator factors as (x + 3)(x - 2)....

which cancels the denominator....

which kills the removable discontinuity...

It remains to prove the limit of x+3 tends to 5 as x tends to 2.

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The "precise" rigorous classic definition:

For every e > 0 there exists d>0 such that

L - e < f(x) < L + e when t-d < x < t+d

and d is a function of e.

Here L=5 and t-2 for function f(x)=x+3

Working backwards:

5 - e < x+3 < 5 + e

2-e < x < 2+e

This suggests we let d=e>0

Then when

2-e < x < 2+e

It MUST follow that

5-e < x+3=f(x) < 5+e

Therefore the function is contained in

the neighborhood of 5 when x is in the

neighborhood of 2.

[end of proof]