Question

a. \(\lim _{x \rightarrow 0} \frac{x-3}{x^{4}-9 x^{2}}\) b. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{4}-9 x^{2}}\) c. \(\lim _{x \rightarrow-3} \frac{x-3}{x^{4}-9 x^{2}}\)

Short answer

Answer: a. \(\lim _{x \rightarrow 0} \frac{x-3}{x^{4}-9 x^{2}} = \frac{1}{27}\) b. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{4}-9 x^{2}} = \frac{1}{108}\) c. \(\lim _{x \rightarrow -3} \frac{x-3}{x^{4}-9 x^{2}}\) does not exist.

Step-by-step solution

Factor the denominator

Factor the denominator using the difference of squares formula: \(x^{4}-9 x^{2} = (x^{2} - 9)(x^{2} + 9)\). Now, the expression is \(\lim _{x \rightarrow 0} \frac{x-3}{(x^{2} - 9)(x^{2} + 9)}\)

Evaluate the limit

Since we have no common factors between the numerator and any of the two factors in the denominator, we can directly plug in the limit value x=0 to the expression: \(\frac{0 - 3}{(0^2 - 9)(0^2 + 9)} = \frac{-3}{(-9)(9)} = \frac{1}{27}\) Answer for part a: \(\lim _{x \rightarrow 0} \frac{x-3}{x^{4}-9 x^{2}} = \frac{1}{27}\) b. \(\lim _{x \rightarrow 3} \frac{x-3}{x^{4}-9 x^{2}}\)

Factor the denominator

The denominator is already factored from part a: \((x^{2} - 9)(x^{2} + 9)\). Now, the expression is \(\lim _{x \rightarrow 3} \frac{x-3}{(x^{2} - 9)(x^{2} + 9)}\)

Cancel out the common factor

Factor the expression \((x-3)\) out of \((x^{2} - 9)\): \((x^{2} - 9) = (x + 3)(x - 3)\). So, the expression after canceling the common factor \((x-3)\) is: \(\lim _{x \rightarrow 3} \frac{1}{(x + 3)(x^{2} + 9)}\)

Evaluate the limit

Plug in the limit value x=3 into the simplified expression: \(\frac{1}{(3 + 3)(3^{2} + 9)} = \frac{1}{(6)(18)} = \frac{1}{108}\) Answer for part b: \(\lim _{x \rightarrow 3} \frac{x-3}{x^{4}-9 x^{2}} = \frac{1}{108}\) c. \(\lim _{x \rightarrow -3} \frac{x-3}{x^{4}-9 x^{2}}\)

Factor the denominator

The denominator is already factored from part a: \((x^{2} - 9)(x^{2} + 9)\). Now, the expression is \(\lim _{x \rightarrow -3} \frac{x-3}{(x^{2} - 9)(x^{2} + 9)}\)

Evaluate the limit

Since we have no common factors between the numerator and any of the two factors in the denominator, we can directly plug in the limit value x=-3 into the expression: \(\frac{-3 - 3}{((-3)^{2} - 9)((-3)^{2} + 9)} = \frac{-6}{(0)(18)}\) Since the denominator is 0, the limit does not exist. Answer for part c: \(\lim _{x \rightarrow -3} \frac{x-3}{x^{4}-9 x^{2}}\) does not exist.

Key concepts

Limit Evaluation

Understanding how to evaluate limits in calculus can be challenging, but with the right approach, it becomes manageable. In essence, evaluating a limit involves finding the value that a function approaches as the input gets arbitrarily close to a certain point. For example, when we see \(\lim _{x \rightarrow c} f(x)\), our task is to determine what value \(f(x)\) approaches as \(x\) gets closer and closer to \(c\).

If the function is continuous at \(c\), simply plugging in the value of \(c\) into the function will give us the limit. If the function is not continuous, we may need to manipulate the expression to resolve any indeterminate form or to factor out common terms. In the given exercise, we've effectively evaluated limits by direct substitution when possible and by simplifying the expression to remove common factors when necessary. It's important to understand both when direct substitution works and when additional algebraic manipulation is required to correctly evaluate limits.

Difference of Squares

The difference of squares is a fundamental algebraic technique used to factor expressions of the form \(A^2 - B^2\), which can be rewritten as \(A + B)(A - B)\). This technique is particularly useful in calculus when simplifying expressions before evaluating limits. In the step-by-step solution provided, the denominator \(x^{4}-9 x^{2}\) was factored using the difference of squares formula into \(x^{2} - 9)(x^{2} + 9)\).

Recognizing when and how to use the difference of squares allows us to simplify complex expressions and often leads to canceling out common terms in the numerator and denominator, thereby revealing the behavior of a function as it approaches a particular value. It's a powerful tool that is used not only for limit evaluation but also in other areas of mathematics such as solving quadratic equations and integrating rational functions.

Indeterminate Forms

Indeterminate forms arise in calculus when the limit of a function yields an undefined expression such as \(0/0\) or \(\infty/\infty\). These forms don't have a clear value until further analysis is done. Taking our exercise as an example, part c presents an indeterminate form when \(x\) approaches -3. The denominator becomes zero, which makes the expression undefined in its current state.

To resolve indeterminate forms, we can use algebraic manipulation like factoring to cancel out common terms or apply L'Hôpital's Rule if conditions are met. The key is to transform the expression into one where the limit can be directly evaluated or indeterminate form no longer exists. Understanding indeterminate forms is essential because they often signal that additional work is needed to find a limit, and they remind us not to jump to conclusions about a function's behavior based on the initial, undefined expression.