+322 Prove that the lim x→3 3x+4=13 using the formal definition of a limit.
+6250 \(\forall \epsilon > 0 \text{ show }\exists \delta \ni |x-3| < \delta \Rightarrow |3x+4-13|< \epsilon\)
\(|3x+4-13| < \epsilon \Rightarrow\\ |3x-9| < \epsilon \Rightarrow \\ -\epsilon < 3x-9 < \epsilon \Rightarrow \\ 9-\epsilon < 3x < 9+\epsilon \Rightarrow \\ 3-\dfrac{\epsilon}{3} < x < 3+ \dfrac{\epsilon}{3}\\ |x-3|< \dfrac{\epsilon}{3}\)
\(\text{so choose }\delta = \dfrac{\epsilon}{3} \\ \text{and working everything backwards you end up with}\\ |3x+4-13| < \epsilon\\ \text{as desired}\)
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+80 Can't you just plug in the value 3 for x?
What do you mean by formal definition of a limit?
+6250 \(\forall \epsilon > 0 \text{ show }\exists \delta \ni |x-3| < \delta \Rightarrow |3x+4-13|< \epsilon\)
\(|3x+4-13| < \epsilon \Rightarrow\\ |3x-9| < \epsilon \Rightarrow \\ -\epsilon < 3x-9 < \epsilon \Rightarrow \\ 9-\epsilon < 3x < 9+\epsilon \Rightarrow \\ 3-\dfrac{\epsilon}{3} < x < 3+ \dfrac{\epsilon}{3}\\ |x-3|< \dfrac{\epsilon}{3}\)
\(\text{so choose }\delta = \dfrac{\epsilon}{3} \\ \text{and working everything backwards you end up with}\\ |3x+4-13| < \epsilon\\ \text{as desired}\)
