Question

19-32 Prove the statement using the \(\varepsilon \), \(\delta \) definition of a limit.

20. \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{5}}} \left( {\frac{{\bf{3}}}{{\bf{2}}}x - \frac{{\bf{1}}}{{\bf{2}}}} \right) = {\bf{7}}\)

Short answer

The given limit is true.

Step-by-step solution

Step 1:Assume the values of \(\varepsilon \) and \(\delta \)

Let \(\varepsilon > 0\) and \(\delta > 0\) such that, if \(0 < \left| {x - 5} \right| < \delta \), thenit can be represented as;

\(\begin{aligned}\left| {\left( {\frac{3}{2}x - \frac{1}{2}} \right) - 7} \right| < \varepsilon \\\left| {\frac{3}{2}x - \frac{{15}}{2}} \right| < \varepsilon \end{aligned}\)

Solve the inequality in step 1

Theinequality\(\left| {\frac{3}{2}x - \frac{{15}}{2}} \right| < \varepsilon \)can be solved as:

\(\begin{aligned}\left| {\frac{3}{2}x - \frac{{15}}{2}} \right| < \varepsilon \\\frac{3}{2}\left| {x - 5} \right| < \varepsilon \\\left| {x - 5} \right| < \frac{2}{3}\varepsilon \end{aligned}\)

Consider \(\delta = \frac{2}{3}\varepsilon \), then from the inequality \(0 < \left| {x - 5} \right| < \delta \):

\(\left| {\left( {\frac{3}{2}x - \frac{1}{2}} \right) - 7} \right| < \varepsilon \)

Thus, the value of the expression\(\mathop {\lim }\limits_{x \to 5} \left( {\frac{3}{2}x - \frac{1}{2}} \right) = 7\).