How do you prove that the limit of #(x^2+3x)=18# as x approaches 3 using the epsilon delta proof?
You can't. The limit is not
graph{x^2 + 3 x [10, 10, 5, 5]}
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To prove that the limit of (x^2+3x) as x approaches 3 is 18 using the epsilondelta 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+6x+3 < deltax+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 epsilondelta proof, demonstrating that the limit of (x^2+3x) as x approaches 3 is indeed 18.
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To prove the limit of ( (x^2 + 3x) ) as ( x ) approaches 3 is 18 using the epsilondelta proof, we follow these steps:

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 ).

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

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

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

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

Since ( (x + 6)(x + 3) < \delta^2 = \varepsilon^2 ), we conclude that the limit is 18.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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