Understanding Continuity of Functions

A function \(f(x)\) is considered continuous at a point \(x=a\) if the following conditions are met:

• The function value at that point, \(f(a)\), is defined.
• The limit of the function as \(x\) approaches \(a\), \(\lim_{x \to a} f(x)\), exists.
• The limit of the function at \(a\) is equal to the function value at \(a\), i.e., \(\lim_{x \to a} f(x) = f(a)\).

In this problem, we are given a function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2} - 2{\rm{x}} - 3}}\) for \({\rm{x}} \ne 3\). We are also told that the function is continuous at \({\rm{x}} = 3\). For the function to be continuous at \({\rm{x}} = 3\), the value of \({\rm{f}}(3)\) must be equal to the limit of the function as \({\rm{x}}\) approaches 3. That is, \({\rm{f}}(3) = \lim_{x \to 3} f(x)\).

Calculating the Limit for Continuity

We need to evaluate the limit \(\lim_{x \to 3} \frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2} - 2{\rm{x}} - 3}}\). If we directly substitute \({\rm{x}} = 3\) into the expression, we get:

\(\frac{{3^2 - 9}}{{3^2 - 2(3) - 3}} = \frac{{9 - 9}}{{9 - 6 - 3}} = \frac{0}{0}\)

This is an indeterminate form, which suggests that we can simplify the expression by factoring the numerator and the denominator.

Factoring the Numerator and Denominator

The numerator is a difference of squares:

\({{\rm{x}}^2} - 9 = {{\rm{x}}^2} - 3^2 = ({\rm{x}} - 3)({\rm{x}} + 3)\)

The denominator is a quadratic expression \({{\rm{x}}^2} - 2{\rm{x}} - 3\). We look for two numbers that multiply to -3 and add up to -2. These numbers are -3 and 1. So, the denominator factors as:

\({{\rm{x}}^2} - 2{\rm{x}} - 3 = ({\rm{x}} - 3)({\rm{x}} + 1)\)

Simplifying the Function Expression

Now we can rewrite the function for \({\rm{x}} \ne 3\):

\({\rm{f}}({\rm{x}}) = \frac{({\rm{x}} - 3)({\rm{x}} + 3)}{({\rm{x}} - 3)({\rm{x}} + 1)}\)

Since we are considering the limit as \({\rm{x}}\) approaches 3, \({\rm{x}} \ne 3\). Therefore, \({\rm{x}} - 3 \ne 0\), and we can cancel the \(({\rm{x}} - 3)\) term from the numerator and the denominator:

\({\rm{f}}({\rm{x}}) = \frac{{\rm{x}} + 3}{{\rm{x}} + 1} \quad \text{for } {\rm{x}} \ne 3\)

Evaluating the Limit

Now we can evaluate the limit of the simplified expression as \({\rm{x}}\) approaches 3:

\(\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{{\rm{x}} + 3}{{\rm{x}} + 1}\)

Substitute \({\rm{x}} = 3\) into the simplified expression:

\(\lim_{x \to 3} f(x) = \frac{3 + 3}{3 + 1} = \frac{6}{4} = 1.5\)

Determining the Value of f(3)

For the function to be continuous at \({\rm{x}} = 3\), we must have \({\rm{f}}(3) = \lim_{x \to 3} f(x)\). From our calculation, the limit is 1.5. Therefore, \({\rm{f}}(3)\) must be 1.5.

\({\rm{f}}(3) = 1.5\)

Comparing this value with the given options, we find that \({\rm{f}}(3) = 1.5\) is one of the options.

Summary of Steps

• Understand that continuity at a point means the function value equals the limit at that point.
• Recognize the need to calculate the limit \(\lim_{x \to 3} f(x)\).
• Identify the indeterminate form upon direct substitution.
• Factor the numerator and denominator of the rational function.
• Cancel the common factor \((x-3)\) since \(x \ne 3\) in the limit process.
• Evaluate the limit of the simplified expression.
• Equate the limit value to \(f(3)\) for continuity.

Revision Table: Continuity and Limits

Additional Information: Piecewise Functions and Continuity

Functions are sometimes defined differently for different intervals or specific points. Such functions are called piecewise functions. In this problem, although not explicitly written in piecewise form, the description implies a piecewise function:

\({\rm{f}}\left( {\rm{x}} \right) = \begin{cases} \frac{{{{\rm{x}}^2} - 9}}{{{{\rm{x}}^2} - 2{\rm{x}} - 3}} & \text{if } {\rm{x}} \ne 3 \\ k & \text{if } {\rm{x}} = 3 \end{cases}\)

Here, \(k\) is an unknown value that represents \({\rm{f}}(3)\). The condition of continuity at \({\rm{x}} = 3\) allows us to find this value \(k\).

For continuity at the point where the definition changes (in this case, \({\rm{x}} = 3\)), the value of the function at that point must smoothly connect with the values around it. This 'smooth connection' is mathematically captured by the limit. The function value at the point must be exactly equal to the limit of the function as \({\rm{x}}\) approaches that point.

If the limit \(\lim_{x \to a} f(x)\) exists but is not equal to \(f(a)\) (or if \(f(a)\) is not defined), the function has a removable discontinuity at \(x=a\). By redefining \(f(a)\) to be equal to the limit, we can 'remove' the discontinuity and make the function continuous at that point, as is the case in this problem where we determined the necessary value for \(f(3)\) to ensure continuity.