Question

Other techniques Evaluate the following limits, where a and \(b\) are fixed real numbers. $$\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3}$$

Short answer

Answer: The limit of the function is 4.

Step-by-step solution

Direct Substitution

Attempt to substitute x=3 directly into the given function: $$\frac{(3)^{2}-2(3)-3}{(3)-3}$$ This, however, results in a division by zero (0/0), so we need to find another way to solve this limit.

Simplify the Function

We can try to factor the numerator to see if we can cancel some terms with the denominator. Factor the quadratic expression in the numerator: \(x^2 - 2x - 3 = (x-3)(x+1)\). This gives: $$\frac{(x-3)(x+1)}{x-3}$$

Cancel Common Factors

Clearly, the \((x-3)\) terms can be canceled from the numerator and denominator: $$\frac{(x-3)(x+1)}{x-3}=\frac{\cancel{ (x-3)}(x+1)}{\cancel{x-3}}=x+1$$

Calculate the Limit

Now that the function has been simplified, we can calculate the limit as x approaches 3 by using direct substitution: $$\lim _{x \rightarrow 3} (x+1) = (3)+1$$

Find the Answer

The final result of the limit is $$\lim _{x \rightarrow 3} \frac{x^{2}-2 x-3}{x-3} = \lim _{x \rightarrow 3} (x+1) = (3)+1 = 4$$

Key concepts

Direct Substitution

Direct substitution is often the first technique we use when evaluating limits. The idea is simple: substitute the value the variable is approaching directly into the function. If the function can be solved without issues, you get the limit right away.

However, when substituting here, we got \[\lim _{x \rightarrow 3} \frac{(3)^{2}-2(3)-3}{3-3} = \frac{0}{0},\] which is undefined because division by zero isn’t allowed. This is what's known as an "indeterminate form." Thus, while direct substitution is a quick method, it doesn't always work right away. In such cases, we need other techniques like factoring or simplification to proceed.

Factoring

Factoring is a valuable tool in simplifying algebraic expressions, especially when dealing with limits. In this problem, we had a quadratic expression in the numerator: \[x^2 - 2x - 3.\]Factoring transforms it into its simplest terms using multiplication, making it possible to "see" and eliminate problematic terms.

We noticed that \[x^2 - 2x - 3 = (x-3)(x+1).\]This factorization strategy highlights the components that might cancel out the zero in the denominator, allowing us to simplify the function. It's a stepping stone for the next critical process: cancelling common factors.

Cancel Common Factors

Once the expression is factored, cancelling common factors becomes straightforward. Here, the expression \[\frac{(x-3)(x+1)}{x-3}\]allowed us to cancel the common factor \((x-3)\) from the numerator and the denominator. Once canceled, you are left with a simpler expression \[x + 1.\]

This process requires a good understanding of algebraic principles, ensuring that only identical factors that are common between the numerator and denominator are removed. By simplifying this way, we make the function easier to handle and remove the indeterminate form problem.

Simplification Techniques

Simplification techniques are integral for handling complex limit problems. After canceling common factors, you simplify the expression to the form where direct substitution becomes possible again.

In our example, after canceling \((x-3)\), the function simplified to \[x+1.\]This now allows direct substitution to find the limit as \(x\) approaches 3: \[\lim _{x \rightarrow 3} (x+1) = 3 + 1 = 4.\]

Simplification removes complexities and avoids undefined forms, ensuring a smooth path to solving the limit.