Question

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

19.\(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 3x}}{{{x^2} - 9}}\),

\(x = 3.1,\,3.05,\,\,3.01,\,\,3.001,\,\,3.0001,\,\,2.9,\,\,2.95,\,\,2.99,\,\,2.999,2.9999\)

Short answer

The value of the limit is,\(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 3x}}{{{x^2} - 9}} = \frac{1}{2}\).

Step-by-step solution

Intuitive Definition of a Limit

Let \(f\left( x \right)\) be defined when \(x\) is near the number \(a\).(Which means that \(f\) is defined on some open interval that contains \(a\), except possibly at \(a\) itself.) Write it as

\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\)

The limit of \(f\left( x \right)\) as \(x\) approaches \(a\) is equal to \(L\). When the values of \(f\left( x \right)\) arbitrarily close to \(L\) by restricting \(x\) to be sufficiently close to \(a\) butnot equal to\(a\).

Compare the definition 1 and 2 we obtain the following is true.

\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\), if and only if \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L\), and \(\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = L\).

Step 2:Evaluate the function at the given numbers

Let \(f\left( x \right) = \frac{{{x^2} - 3x}}{{{x^2} - 9}}\).

Evaluate the function at \(x = 3.1\) as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{x \to 3.1} \frac{{{x^2} - 3x}}{{{x^2} - 9}} &=\frac{{{{\left( {3.1} \right)}^2} - 3\left( {3.1} \right)}}{{{{\left( {3.1} \right)}^2} - 9}}\\ &=\frac{{9.61 - 9.3}}{{9.61 - 9}}\\&=\frac{{0.31}}{{0.61}}\\ &= 0.508197\end{aligned}\)

The other values are obtained in a similar way listed in the table as shown below:

\(x\)	\(f\left( x \right)\)
\(\begin{aligned}{l}3.1\\3.05\\3.01\\3.001\\3.0001\\2.9\\2.95\\2.99\\2.999\\2.9999\end{aligned}\)	\(\begin{array}{l}0.508197\\0.504132\\0.500183\\0.500083\\0.500008\\0.491525\\0.495795\\0.499165\\0.499917\\0.499992\end{array}\)

It is observed from the table that left, and right limits are the same and therefore, \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 3x}}{{{x^2} - 9}} = 0.5 \approx \frac{1}{2}\).

Thus, the value of the limit is \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 3x}}{{{x^2} - 9}} = \frac{1}{2}\).