In Example 1, we claimed that \(\lim\limits_{x\rightarrow 2}(2x+5) =9\).

How close must \(x\) be to 2, so that \(f(x) =2x+5\) is within 0.1 of 9?

How close must \(x\) be to 2, so that \(f(x) =2x+5\) is within 0.05 of 9?

On the number line, the distance between two points with coordinates \({a}\) and \(b\) is \(\vert{a-b}\vert\).

Solution (a) The function \(f(x)=2x+5\) is within 0.1 of 9, if the distance between \(f(x)\) and 9 is less than 0.1 unit. That is, if \(\vert f(x)-9\vert\leq 0.1\). \[ \begin{eqnarray*} \vert (2x+5) -9\vert &\leq &0.1 \\ \vert 2x-4\vert &\leq &0.1 \\ \vert 2( x-2) \vert &\leq &0.1 \\ \vert x-2\vert &\leq &\dfrac{0.1}{2}=0.05 \\ -0.05 &\leq &x-2\leq 0.05 \\ 1.95 &\leq &x\leq 2.05 \end{eqnarray*} \]

Inequalities involving absolute values are discussed in Appendix A.1, p. A-7.

So, if \(1.95\leq x\leq 2.05\), then \(f(x)\) will be within 0.1 of 9.

(b)The function \(f(x)=2x+5\) is within 0.05 of 9 if \(\vert f(x) -9\vert \leq 0.05\). That is, \[ \begin{eqnarray*} \vert ( 2x+5) -9\vert &\leq &0.05 \\ \vert 2x-4\vert &\leq &0.05 \\ \vert x-2\vert &\leq &\dfrac{0.05}{2}=0.025 \end{eqnarray*} \]

\(f(x)=2x+5\)

So, if \(1.975\leq x\leq 2.025\), then \(f(x)\) will be within 0.05 of 9.