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# Calculus

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Prove that the lim x→3   3x+4=13  using the formal definition of a limit.

Feb 5, 2019

#2
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$$\forall \epsilon > 0 \text{ show }\exists \delta \ni |x-3| < \delta \Rightarrow |3x+4-13|< \epsilon$$

$$|3x+4-13| < \epsilon \Rightarrow\\ |3x-9| < \epsilon \Rightarrow \\ -\epsilon < 3x-9 < \epsilon \Rightarrow \\ 9-\epsilon < 3x < 9+\epsilon \Rightarrow \\ 3-\dfrac{\epsilon}{3} < x < 3+ \dfrac{\epsilon}{3}\\ |x-3|< \dfrac{\epsilon}{3}$$

$$\text{so choose }\delta = \dfrac{\epsilon}{3} \\ \text{and working everything backwards you end up with}\\ |3x+4-13| < \epsilon\\ \text{as desired}$$

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Feb 5, 2019

#1
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Can't you just plug in the value 3 for x?

What do you mean by formal definition of a limit?

Feb 5, 2019
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I wish it was that easy

Ruublrr  Feb 6, 2019
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$$\forall \epsilon > 0 \text{ show }\exists \delta \ni |x-3| < \delta \Rightarrow |3x+4-13|< \epsilon$$

$$|3x+4-13| < \epsilon \Rightarrow\\ |3x-9| < \epsilon \Rightarrow \\ -\epsilon < 3x-9 < \epsilon \Rightarrow \\ 9-\epsilon < 3x < 9+\epsilon \Rightarrow \\ 3-\dfrac{\epsilon}{3} < x < 3+ \dfrac{\epsilon}{3}\\ |x-3|< \dfrac{\epsilon}{3}$$

$$\text{so choose }\delta = \dfrac{\epsilon}{3} \\ \text{and working everything backwards you end up with}\\ |3x+4-13| < \epsilon\\ \text{as desired}$$

Rom Feb 5, 2019
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I'm sure you are correct Rom but I am with itsyaboi.

Talking about needing to go the long way around!

Melody  Feb 6, 2019
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well he did ask to use the formal definition of a limit

Rom  Feb 6, 2019
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Yes i understand that.

Sometimes I think mathematicians (not you) purposely design things to be complicated when it does not seem remotely necessary.

Melody  Feb 6, 2019