Question

In Exercises \(29-32,\) find the limit \(L .\) Then find \(\delta>0\) such that \(|f(x)-L|<0.01\) whenever \(0<|x-c|<\delta\). $$ \lim _{x \rightarrow 2}(3 x+2) $$

Short answer

The limit \(L = 8\) and \(\delta = 0.00333\).

Step-by-step solution

Find the limit

To determine the limit of the function \(f(x) = 3x + 2\) as \(x\) approaches 2 (i.e. \(L = \lim_{x->2} (3x + 2)\)), you substitute \(x = 2\) into the function. This gives \(L = 3(2) + 2 = 8\).

Find \(\delta\)

With \(L = 8\), we need to find \(\delta > 0\) such that \(|f(x)-L| < 0.01\) whenever \(0 < |x-2| < \delta\). This inequality simplifies to -0.01 < (3x+2) - 8 < 0.01, and further simplifies to 7.99 < 3x < 8.01. Solving these for \(x\) gives \(\frac{7.99}{3} < x < \frac{8.01}{3}\). Note that \(x=2\) lies in this interval. Lastly, we choose \(\delta\) to be the smaller of the two distances between \(x=2\) and the two ends of this interval. Calculating these distances gives \(\delta = min(2 - \frac{7.99}{3}, \frac{8.01}{3} - 2)\). Simplifying gives \(\delta = min(0.00333, 0.00333)\), so we find \(\delta = 0.00333\).