How do you prove the statement lim as x approaches -3 for #(x^2+3x)# using the epsilon and delta definition?

Answer 1

See the explanation.

I assume that the statement we want to prove is #lim_(xrarr-3)(x^2+3x)=0#

Initial Work

The #epsilon, delta# proofs for quadratic functions are more involved than those for linear functions. Most of us need to be shown this method, we don't come up with it by ourselves.
We begin (as we do for linear functions) by examining the difference between #f(x)# and #L#. Keep in mind that, for his question #abs(x-a) = abs(x-(-3)) = abs(x+3)
Now, #abs((x^2+3x)-0) = abs(x(x+3)) = absx abs(x+3)#
We can control, through the choice of #delta#, the size of #abs(x+3)#.
If #abs(x+3)# is small, then #x# is close to #-3#. We'll put a bound on #absx# by choosing #delta# so that is is less than some positive number we select now. #1# is easy to work with, so we'll make sure that #delta <= 1#. This will assure us that when #abs(x-(-3)) < delta <= 1#, we have
#x# is within #1# unit of #-3#,
So, #-3-1 < x < -3+1#
I.e. #-4 < x , -2#,
so #absx < 4#.
If we also make sure that #delta <= epsilon/4#, the we will have:
When #abs(x+3) < delta#, then
#absx abs(x+3) < 4abs(x+3) < 4(epsilon/4) = epsilon#

Proof

Given #epsilon > 0#, choose #delta = min{1, epsilon/4}#
If #0 < abs(x+3) < delta#, then
#abs((x^2+3x)-0) = absx abs(x+3) < 4abs(x+3) <= 4(epsilon/4)=epsilon#
That is, #abs((x^2+3x)-0) < epsilon#.
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Answer 2

To prove the statement lim as x approaches -3 for (x^2+3x), we need to show that for any given epsilon greater than zero, there exists a delta greater than zero such that if the distance between x and -3 is less than delta, then the distance between (x^2+3x) and the limit L is less than epsilon.

Let's proceed with the proof:

Given the function f(x) = x^2 + 3x, we want to find the limit as x approaches -3.

We start by assuming an arbitrary epsilon > 0.

We need to find a delta > 0 such that if |x - (-3)| < delta, then |(x^2 + 3x) - L| < epsilon.

Let's simplify the expression |(x^2 + 3x) - L|:

|(x^2 + 3x) - L| = |x^2 + 3x + 3|.

Now, we can manipulate the expression |x^2 + 3x + 3| to find a suitable delta.

We can rewrite |x^2 + 3x + 3| as |(x + 3)(x)|.

Since we are interested in the behavior of the function as x approaches -3, we can assume that |x + 3| < 1 (or any other positive value) to simplify the expression further.

Therefore, |(x + 3)(x)| < |1 * x| = |x|.

Now, we want to find a delta such that if |x - (-3)| < delta, then |x| < epsilon.

Since |x| < epsilon, we can choose delta = epsilon.

Thus, for any given epsilon > 0, if |x - (-3)| < delta = epsilon, then |(x^2 + 3x) - L| < epsilon.

Therefore, we have proven the statement using the epsilon and delta definition.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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