Question

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest such δ.) What limit does this prove?

Answer 1

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$, we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$. Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$.

Then, if $|x-5|$, then $|f(x)-14|$. So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$. The maximum value of delta is $\frac{\varepsilon}{3}$.

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$. Note that f(5)=14, so this proves that f is continous at x=5.