Question

a. \(\lim _{x \rightarrow 0} \frac{x-2}{x^{5}-4 x^{3}}\) b. \(\lim _{x \rightarrow 2} \frac{x-2}{x^{5}-4 x^{3}}\) c. \(\lim _{x \rightarrow-2} \frac{x-2}{x^{5}-4 x^{3}}\)

Short answer

Question: Find the following limits: a. \(\lim _{x \rightarrow 0} \frac{x-2}{x^{5}-4 x^{3}}\) b. \(\lim _{x \rightarrow 2} \frac{x-2}{x^{5}-4 x^{3}}\) c. \(\lim _{x \rightarrow -2} \frac{x-2}{x^{5}-4 x^{3}}\) Answer: a. The limit does not exist. b. \(\frac{1}{64}\) c. \(\frac{1}{16}\)

Step-by-step solution

Directly substitute the value of x

Substitute x=0 into the expression: \(\frac{0-2}{0^{5}-4(0)^{3}} =\frac{-2}{0}\). This results in an indeterminate form, so we need to find another way to evaluate this limit.

Simplify the expression

Factor out an \(x^3\) from the denominator: \(\lim _{x \rightarrow 0}\frac{x-2}{x^3(x^2-4)}\). Now we can try to cancel common factors from the numerator and denominator.

Evaluate the limit

None of the factors in the denominator can be canceled with the numerator. So, we can directly find the limit by letting x approach 0, and the expression approaches: \(\frac{-2}{0(0^2-4)} =\boxed{\frac{-2}{0}}\) which does not exist. b. \(\lim _{x \rightarrow 2} \frac{x-2}{x^{5}-4 x^{3}}\)

Directly substitute the value of x

Substitute x=2 into the expression: \(\frac{2-2}{2^{5}-4(2)^{3}} =\frac{0}{32-32}\). This results in an indeterminate form, so we need to find another way to evaluate this limit.

Simplify the expression

Factor out x-2 from the denominator, but we can notice that the denominator itself is not completely factorizable: \(\lim _{x \rightarrow 2}\frac{x-2}{x^3(x^2-4)}\). Regardless of this fact, we can simply cancel the \(x-2\) term from the numerator and the denominator.

Evaluate the limit

Now the expression becomes \(\lim _{x \rightarrow 2}\frac{1}{x^3(x+2)}\). Substitute x=2, and we get \(\lim _{x \rightarrow 2}\frac{1}{2^3(2+2)}=\boxed{\frac{1}{64}}\) c. \(\lim _{x \rightarrow -2} \frac{x-2}{x^{5}-4 x^{3}}\)

Directly substitute the value of x

Substitute x=-2 into the expression: \(\frac{-2-2}{(-2)^{5}-4(-2)^{3}} =\frac{-4}{-32-32}\). This does not result in an indeterminate form, so we can directly evaluate the limit.

Evaluate the limit

The expression becomes \(\lim _{x \rightarrow -2}\frac{x-2}{x^5-4x^3}= \boxed{\frac{-4}{-64}} = \boxed{\frac{1}{16}}\).

Key concepts

Indeterminate Forms

When working with limits in calculus, one may often come across what are known as 'indeterminate forms.' These forms appear when the substitution of a point into a limit expression results in an undefined or ambiguous value, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In the given exercise, we encounter indeterminate forms in parts (a) and (b), where direct substitution leads to \(\frac{-2}{0}\) and \(\frac{0}{0}\), respectively.

Indeterminate forms require additional methods to resolve so that the limit can be evaluated. For example, one might use algebraic manipulation, L'Hôpital's Rule, or look into the behavior of the functions around the point in question. It's important not to misconstrue these expressions as simply being zero or non-existent; they signal that further investigation is needed for a correct evaluation of the limit.

Factorization in Limits

Factorization is a powerful algebraic tool that can transform indeterminate forms into something more workable when evaluating limits. It involves breaking down polynomials into simpler products that may reveal common factors in the numerator and denominator.

In part (b) of our problem, factorization allows us to cancel out the \(x-2\) factor from both the numerator and denominator once we recognize the indeterminate form. This step simplifies the expression greatly, and what follows is a straightforward substitution. By factoring, we reduce complex expressions into simpler ones which can often lead to an easily computable limit. Students should be encouraged to look for opportunities to factorize whenever they encounter complex polynomials within limit problems.

Limits of Rational Functions

Rational functions, which are quotients of two polynomials, can lead to limits that are indeterminate forms as seen in the exercise. However, after applying factorization, these types of limits can often be resolved into definite values.

In parts (b) and (c) of the exercise, we see different approaches to solving limits involving rational functions. In (b), simplification through factorization is key, while in (c), the limit can be immediately determined without needing to resolve an indeterminate form. When faced with a rational function, students should first assess if direct substitution works. If it leads to an indeterminate form, factorization or other techniques should be applied to find the limit. Being proficient in these methods is essential for students to tackle limits of rational functions effectively.