Question

Evaluate the limit \(\lim _{x \rightarrow a} \frac{x-a}{x^{2}-a^{2}}, \quad a \neq 0\)

Short answer

The limit is \(\frac{1}{2a}\).

Step-by-step solution

Identify the Function and Indeterminate Form

We need to evaluate the limit \(\lim _{x \rightarrow a} \frac{x-a}{x^{2}-a^{2}}\). Substituting \(x = a\) results in \(\frac{0}{0}\), which is an indeterminate form.

Factor the Denominator

Notice the denominator can be factored using the difference of squares: \(x^2 - a^2 = (x-a)(x+a)\).

Simplify the Expression

Rewrite the original limit as \(\lim _{x \rightarrow a} \frac{x-a}{(x-a)(x+a)}\). Cancel the common \((x-a)\) terms: \(\lim _{x \rightarrow a} \frac{1}{x+a}\), valid for \(x eq a\).

Substitute the Limit

Substitute \(x = a\) into the simplified expression: \(\frac{1}{a+a} = \frac{1}{2a}\).

Conclusion

Thus, the limit evaluates to \(\frac{1}{2a}\).

Key concepts

Indeterminate Forms

In calculus, indeterminate forms occur when substituting a value into a function results in an ambiguous expression, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not convey enough information about the behavior of the function at that point. Instead, they serve as a signal that more manipulation is necessary to find the limit.

When you encounter an indeterminate form, it means you need to apply algebraic techniques to transform the expression into a determinate form, one that yields a meaningful limit. One common scenario leading to the \(\frac{0}{0}\) form involves polynomial functions, as we see in our original exercise.

By recognizing the indeterminate form at \(x = a\), you identify the need for further simplification or a change of perspective to accurately evaluate the limit. Often, this involves factoring or canceling terms that make the form indeterminate.

Factoring Polynomials

Factoring is a crucial algebraic process that breaks down a polynomial into simpler, more manageable pieces called factors. In our exercise, the expression \(x^2 - a^2\) in the denominator is a perfect candidate for factoring.

Here, we have a special type of polynomial known as the difference of squares, represented as \(x^2 - a^2\). This can always be factored into \((x-a)(x+a)\).

How to Factor the Difference of Squares:

• Recognize the form \((number)^2 - (number)^2\).
• Break it down into \((first number - second number)(first number + second number)\).

By factoring \(x^2 - a^2 = (x-a)(x+a)\), the expression becomes easier to handle and simplify, which is vital for finding limits in calculus. Factoring often reveals cancellations that simplify expressions greatly, as seen when the \((x-a)\) term cancels out in our problem.

Simplifying Expressions

Simplifying expressions is the art of making a problem more tractable by reducing it to its simplest form. In the context of calculus and limits, simplifying the expression means eliminating factors or terms that make direct substitution problematic.

For our original problem, once we factored the denominator into \((x-a)(x+a)\), the next step was to simplify by canceling the common \((x-a)\) factor in the numerator and denominator. After canceling, the expression is transformed into \(\frac{1}{x+a}\).

Key Points in Simplification:

• Look for common terms in the numerator and denominator.
• Cancel these terms to prevent the \(\frac{0}{0}\) form.
• Ensure that any cancellations are mathematically valid across all parts of the expression, except the point of interest (\(x=a\)). This makes the expression easier to solve.

Once simplified, you can substitute to find the limit meaningfully without ambiguity. In our example, substituting \(x = a\) into \(\frac{1}{x+a}\) now accurately gives the limit \(\frac{1}{2a}\), as the expression is no longer problematic.