Question

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow 1}\left(\frac{2}{3} x+9\right) $$

Short answer

The limit is \(L = \frac{29}{3}\). For each value of \(\varepsilon\), a corresponding \(\delta = 3 \varepsilon\) fulfills the epsilon-delta definition.

Step-by-step solution

Find the limit

We substitute the value \(x = 1\) into the function. The limit as \(x\) approaches 1 for the function \(\frac{2}{3}x + 9\) can be calculated as follows: \[ \lim _{x \rightarrow 1}\left(\frac{2}{3} x+9\right) = \frac{2}{3}(1) + 9 = \frac{2}{3} + 9 = \frac{29}{3} \]

Epsilon-Delta Proof

We need to prove that for every positive number \(\varepsilon\) there exists a positive number \(\delta\) such that when the difference between \(x\) and 1 is less than \(\delta\) (but not equal to 0), the difference between \(\frac{2}{3} x + 9\) and \(\frac{29}{3}\) is less than \(\varepsilon\). Let's control the absolute difference: \[ \left| \left(\frac{2}{3}x + 9\right) - \frac{29}{3}\right| = \left|\frac{2}{3}(x - 1)\right| \] We need this to be less than \(\varepsilon\) when \(0<|x-1|< \delta\). Let's select \(\delta = 3 \varepsilon\). If \(0<|x-1|< \delta\), then it holds: \[ |(x-1)|\frac{2}{3}<\varepsilon \] Therefore, for every \(\varepsilon > 0\), we have found a \(\delta\)(specifically, \(\delta = 3\varepsilon\)) which fulfills the epsilon-delta definition of the limit.