How do you prove that the limit of #(x^2+3x)=-18# as x approaches -3 using the epsilon delta proof?

Answer 1

You can't. The limit is not #-18#.

#lim_(xrarr-3)(x^2+3x) = (-3)^2+3(-3) = 9-9=0#

graph{x^2 + 3 x [-10, 10, -5, 5]}

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Answer 2

To prove that the limit of (x^2+3x) as x approaches -3 is -18 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that whenever 0 < |x - (-3)| < delta, then |(x^2+3x) - (-18)| < epsilon.

Let's proceed with the proof:

Given epsilon > 0, we need to find a suitable delta > 0. We start by manipulating the expression |(x^2+3x) - (-18)| < epsilon:

|(x^2+3x) + 18| < epsilon |(x+6)(x+3)| < epsilon

Now, we can observe that if we restrict the value of delta such that 0 < |x - (-3)| < delta, then we can assume that |x + 3| < delta. This allows us to make the following inequality:

|(x+6)(x+3)| = |x+6||x+3| < delta|x+6|

To proceed further, we can choose a value for delta such that delta < 1. This implies that |x + 6| < 7 for all x satisfying 0 < |x - (-3)| < delta.

Therefore, we have:

|(x+6)(x+3)| < 7delta

Now, we can choose delta = epsilon/7. This ensures that whenever 0 < |x - (-3)| < delta, we have:

|(x+6)(x+3)| < 7delta = 7(epsilon/7) = epsilon

Thus, we have shown that for any given epsilon > 0, we can find a suitable delta > 0 (specifically, delta = epsilon/7) such that whenever 0 < |x - (-3)| < delta, then |(x^2+3x) - (-18)| < epsilon. This completes the epsilon-delta proof, demonstrating that the limit of (x^2+3x) as x approaches -3 is indeed -18.

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Answer 3

To prove the limit of ( (x^2 + 3x) ) as ( x ) approaches -3 is -18 using the epsilon-delta proof, we follow these steps:

  1. Begin by stating the definition of the limit: For any ( \varepsilon > 0 ), there exists a ( \delta > 0 ) such that if ( 0 < |x - (-3)| < \delta ), then ( |(x^2 + 3x) - (-18)| < \varepsilon ).

  2. Compute the expression ( |(x^2 + 3x) - (-18)| ) and simplify it to ( |x^2 + 3x + 18| ).

  3. Factorize ( x^2 + 3x + 18 ) as ( |(x + 6)(x + 3)| ).

  4. Notice that as ( x ) approaches -3, ( (x + 3) ) approaches 0.

  5. Set ( \delta = \varepsilon ) and observe that if ( 0 < |x - (-3)| < \delta ), then ( |(x^2 + 3x) - (-18)| = |(x + 6)(x + 3)| < \delta^2 ).

  6. Since ( |(x + 6)(x + 3)| < \delta^2 = \varepsilon^2 ), we conclude that the limit is -18.

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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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