Question

Other techniques Evaluate the following limits, where a and \(b\) are fixed real numbers. $$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$$

Short answer

Answer: 2

Step-by-step solution

Identify the indeterminate form of the function

The function is given as: $$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$$ We notice that both the denominator and numerator approach zero when \(x \rightarrow 1\). This type of indeterminate form is called \(0/0\).

Simplify the function if possible

We want to check if we can simplify the function to avoid the division by zero. The numerator is the difference of squares, which means we can factor it as \((x+1)(x-1)\). This gives us: $$\lim _{x \rightarrow 1} \frac{(x+1)(x-1)}{x-1}$$

Cancel out common factors

We notice that the numerator and denominator share a common factor of \((x-1)\). Hence, we can cancel it out: $$\lim _{x \rightarrow 1} (x+1)$$

Evaluate the limit

Now that the function has been simplified and doesn't have an indeterminate form anymore, we can evaluate the limit by plugging in \(x = 1\): $$\lim _{x \rightarrow 1} (x+1) = (1+1) = 2$$

Write the final answer

The limit of the given function as \(x\) approaches 1 is 2: $$\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1} = 2$$

Key concepts

Indeterminate Form

Many students encounter indeterminate forms when studying limits. An indeterminate form occurs when the result of a limit is not immediately clear. For example, the expression \( \frac{0}{0} \) is an indeterminate form because zero divided by zero does not have a defined value. In the provided exercise, as \( x \) approaches 1, both the numerator \( x^2 - 1 \) and the denominator \( x - 1 \) approach zero. This results in a \( \frac{0}{0} \) indeterminate form. Recognizing indeterminate forms is crucial because it signals the need for further algebraic manipulation to resolve the limit.

Simplifying Functions

Simplifying functions is often the key to evaluating limits that result in indeterminate forms. By simplifying, you can remove the indeterminate form and make the function easier to work with. In our example, the function \( \frac{x^2 - 1}{x - 1} \) seems complex due to the presence of zero in both the numerator and the denominator as \( x \) approaches 1. To simplify, you should look for ways to factor or reduce the expression. This could mean factoring polynomials, canceling terms, or any algebraic steps that transform the expression into a more manageable form. Once simplified, the expression becomes easier to evaluate, as we see with the example.

Factorization

Factorization is a fundamental algebraic technique used to simplify expressions, especially when handling limits. In the given exercise, the numerator \( x^2 - 1 \) is a difference of squares, which can be factored into \( (x+1)(x-1) \). This step is essential because it reveals any common factors that can be canceled with the denominator, simplifying the limit problem. By applying factorization, the original expression \( \frac{x^2 - 1}{x - 1} \) becomes \( \frac{(x+1)(x-1)}{x-1} \). This allows for the cancellation of the \( (x-1) \) factor in both the numerator and the denominator, thus removing the indeterminate form and simplifying the expression to \( x+1 \).

Evaluating Limits

Once you have simplified the expression and removed the indeterminate form, evaluating the limit becomes straightforward. You can substitute the value that \( x \) approaches directly into the simplified function. In our example, after simplification, we have \( \lim_{x \to 1} (x+1) \). Here, the limit is easy to evaluate by substituting \( x = 1 \), resulting in \( 1 + 1 = 2 \). Evaluating limits is like solving a puzzle: once the pieces (in the form of simplification and factorization) are correctly placed, the solution becomes clear. Always ensure the function is defined at the point of evaluation, free from indeterminate forms, to accurately find the limit.