#2**+2 **

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

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

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

.Rom Feb 5, 2019

#1**+1 **

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

What do you mean by formal definition of a limit?

itsyaboi Feb 5, 2019

#2**+2 **

Best Answer

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

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

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

Rom Feb 5, 2019