Question

How do you prove that the limit `f(x)= x^2 + 2x - 5 =3` as x approaches 2 using the formal definition of a limit?

Answer 1

See below

Explanation:

The function is continuous so you can simply say:

`lim_(x to 2) x^2 + 2x - 5 = f(2)= 3`

Answer 2

See below for application of formal definition of limit.

Explanation:

Since the given function is continuous and defined at the limit the value of the function as `x` approaches the limit is simply the value of the function at the limit.
However, the objective seems to be to demonstrate an application of the formal definition.
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Formal Definition of Limit

A function `f(x)` has a limit of `c` as `x` approaches `a`,
typically written as: `lim_(xrarra)f(x) =c`
if and only if
for any value `delta` greater than zero,
`color(white)("XXX")` whenever `0 < abs(x-a) < delta`
then there exists some value `epsilon` (usually dependent on `delta` and sometimes written `epsilon_delta`) such that
`color(white)("XXX")abs(f(x)-c) < epsilon_delta`

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Application of Formal Definition

Suppose `f(x)=x^2+2x-5`
and we wish to show that `lim_(xrarr2)f(x)=3`

For any value `delta` such that `0 < abs(x-2) < delta`
We need to find a value `epsilon_delta` such that
`color(white)("XXX")0 < abs(f(x)-3) < epsilon_delta`
(we can safely ignore the requirement of `0 <`)
and we have specifically:
`color(white)("XXX")abs((x^2+2x-5)-3) < epsilon_delta`

`color(white)("XXX")rarr abs(x^2+2x-8) < epsilon_delta`

`color(white)("XXX")rarr abs((x-2) * (x+4)) < epsilon_delta`

`color(white)("XXX")rarr abs(x-2) * abs(x+4) < epsilon_delta`
`color(white)("XXXXX")`notice the factor of `abs(x-2)` and its relation to the limitation on `delta`

Since we are interested in the domain where `xrarr2`, we an limit `x in (1,3)`
and in this area `abs(x+4) =x+4 < 7`

So
`color(white)("XXX")abs(f(x)-3)=abs(x-2) * abs(x+4) < 7abs(x-2)`
and we know that
`color(white)("XXX")abs(x-2) < delta`
so
`color(white)("XXX")abs(f(x)-3) < delta/7`

If we pick `epsilon_delta = delta/7`
then
`color(white)("XXX")abs(f(x)-3) < epsilon_delta` provided `0 < delta < abs(x-2)`
completing the formal proof of the limit as required.

Answer 3

Given `epsilon > 0` choose `delta = min{1, epsilon/7}`. Note that `delta > 0`.

For every `x` with `0 < abs(x-2) < delta`, we have

`abs(x-2) < 1`, so `-1 < x-2 < 1` and `1 < x < 3`.

This entails that `5 < x+4 < 7` , so that `abs(x+4) < 7`.

Summarizing, for all `x` such that `0 < abs(x-2) < delta`, we have `abs(x+4) < 7`.

Now `abs((x^2+2x-5)-3) = abs(x^2+2x-8) = abs(x+4)abs(x-2)`

And if `abs(x-2) < delta` then `abs(x+4) < 7`, and `abs(x-2) < epsilon/7`.

So, if `abs(x-2) < delta`, then

`abs((x^2+2x-5)-3) = abs(x+4)abs(x-2)`

`< (7)*(epsilon/7) = epsilon`.

We have shown that for any positive `epsilon` there is a positive `delta` such that for any `x` with `0 < abs(x-2) < delta`, we have `abs((x^2+2x-5)-(3)) < epsilon`.

By the definition of limit, `lim_(xrarr2)(x^2+2x-5)= 3`.