Answer:
This proves that f is continous at x=5.
Step-by-step explanation:
Taking f(x) = 3x-1 and [tex]\varepsilon>0[/tex], we want to find a [tex]\delta [/tex] such that [tex]|f(x)-14|<\varepsilon[/tex]
At first, we will assume that this delta exists and we will try to figure out its value.
Suppose that [tex]|x-5|<\delta[/tex]. Then
[tex]|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta[/tex].
Then, if [tex]|x-5|<\delta[/tex], then [tex]|f(x)-14|<3\delta[/tex]. So, in this case, if [tex]3\delta \leq \varepsilon[/tex] we get that [tex]|f(x)-14|<\varepsilon[/tex]. The maximum value of delta is [tex]\frac{\varepsilon}{3}[/tex].
By definition, this procedure proves that [tex]\lim_{x\to 5}f(x) = 14[/tex]. Note that f(5)=14, so this proves that f is continous at x=5.
