Question

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

31. \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 1} \right) = 3\)

Short answer

It is proved that \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 1} \right) = 3\).

Step-by-step solution

Precise Definition of a limit

Consider \(f\) as afunctiondefined on some open intervalthat contains the number \(a\), except possibly at \(a\) itself.

Then, we say that thelimit of \(f\left( x \right)\) as \(x\)approaches \(a\) is \(L\), and it can be written as\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\).

If for all numbers \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that if \(0 < \left| {x - a} \right| < \delta \) then \(\left| {f\left( x \right) - L} \right| < \varepsilon \).

Prove the statement using the definition of a limit

Let \(\varepsilon > 0\)be a given positive number.

We want \(\delta > 0\) such that when\(0 < \left| {x - \left( { - 2} \right)} \right| < \delta \) then \(\left| {\left( {{x^2} - 1} \right) - 3} \right| < \varepsilon \) or by simplifying we require \(\left| {{x^2} - 4} \right| < \varepsilon \) whenever \(0 < \left| {x + 2} \right| < \delta \).

It is observed that when \(\left| {x + 2} \right| < 1\) then, it is represented as:

\(\begin{array}{c} - 1 < x + 2 < 1\\ - 3 < x < - 1\\ - 5 < x - 2 < - 3\\\left| {x - 2} \right| < 5\end{array}\)

Let us choose \(\delta = \min \left\{ {\frac{\varepsilon }{5},1} \right\}\). If \(0 < \left| {x + 2} \right| < \delta \) then \(\left| {x - 2} \right| < 5\) and \(\left| {x + 2} \right| < \frac{\varepsilon }{5}\).

Therefore,

\(\begin{aligned}\left| {\left( {{x^2} - 1} \right) - 3} \right| &= \left| {{x^2} - {2^2}} \right|\\ &= \left| {\left( {x + 2} \right)\left( {x - 2} \right)} \right|\\ &= \left| {x + 2} \right|\left| {x - 2} \right|\\ < \left( {\frac{\varepsilon }{5}} \right)\left( 5 \right)\\ &= \varepsilon \end{aligned}\)

Hence, \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 1} \right) = 3\) according to the definition of a limit.

Thus, it is proved that \(\mathop {\lim }\limits_{x \to - 2} \left( {{x^2} - 1} \right) = 3\).