Question

Prove the statement using the \(\varepsilon \) \(\delta \)definition of a limit.

\(\mathop {\lim }\limits_{x \to - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\)

Short answer

It is proved that\(\mathop {\lim }\limits_{x \to - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\).

Step-by-step solution

Describe the given information

It is required to prove\(\mathop {\lim }\limits_{x \to - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\)by using\(\varepsilon \),\(\delta \)definition.

Prove that \(\mathop {\lim }\limits_{x \to  - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\)

Consider the limit\(\mathop {\lim }\limits_{x \to - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\).

Use \(\varepsilon - \delta \) definition to prove the statement\(\mathop {\lim }\limits_{x \to - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\).

If \(0 < \left| {x - \left( { - 1.5} \right)} \right| < \delta \)then\(0 < \left| {\left( {\frac{{9 - 4{x^2}}}{{3 + 2x}}} \right) - 6} \right| < \varepsilon \).

Simplify the absolute value inequality as follows.

If \(0 < \left| {x + 1.5} \right| < \delta \)then\(0 < \left| {\frac{{\left( {3 + 2x} \right)\left( {3 - 2x} \right)}}{{3 + 2x}} - 6} \right| < \varepsilon \).

If \(0 < \left| {x + 1.5} \right| < \delta \)then\(0 < \left| {\left( {3 - 2x} \right) - 6} \right| < \varepsilon \).

If \(0 < \left| {x + 1.5} \right| < \delta \)then \(0 < \left| { - 3 - 2x} \right| < \varepsilon \).

Factor out \(2\) from the inequality \(0 < \left| { - 3 - 2x} \right| < \varepsilon \)

\(\begin{array}{c}0 < \left| {\left( 2 \right)\left( { - 1} \right)\left( {x + 1.5} \right)} \right| < \varepsilon \\0 < 2\left| {x + 1.5} \right| < \varepsilon \\0 < \left| {x + 1.5} \right| < \frac{\varepsilon }{2}\end{array}\).

If \(0 < \left| {x + 1.5} \right| < \delta \)then\(0 < \left| {x + 1.5} \right| < \frac{\varepsilon }{2}\).

Choose\(\delta = \frac{\varepsilon }{2}\), and then simplify the inequality.

\(\begin{array}{c}0 < 2\left| {x + 1.5} \right| < \varepsilon \\0 < \left| {2x + 3} \right| < \varepsilon \\0 < \left| {\frac{{{{\left( {2x + 3} \right)}^2}}}{{3 + 2x}}} \right| < \varepsilon \\0 < \left| {\frac{{4{x^2} + 12x + 9}}{{3 + 2x}}} \right| < \varepsilon \end{array}\)

Simplify further.

\(\begin{array}{c}0 < \left| {\frac{{ - 4{x^2} - 12x - 9}}{{3 + 2x}}} \right| < \varepsilon \\0 < \left| {\frac{{9 - 4{x^2} - 12x - 18}}{{3 + 2x}}} \right| < \varepsilon \\0 < \left| {\frac{{9 - 4{x^2}}}{{3 + 2x}} - 6} \right| < \varepsilon \end{array}\)

Therefore, it is proved that\(\mathop {\lim }\limits_{x \to - 1.5} \frac{{9 - 4{x^2}}}{{3 + 2x}} = 6\).