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Chapter 2: Problem 28

Find the limits in Exercises \(19-36\). $$ \lim _{v \rightarrow 2} \frac{v^{3}-8}{v^{4}-16} $$

Short Answer

Expert verified
The limit is \( \frac{3}{8} \).

Step by step solution

01

Identify the Indeterminate Form

Plugging in the limit value, \( v = 2 \), into the function \( \frac{v^3 - 8}{v^4 - 16} \) results in \( \frac{8 - 8}{16 - 16} = \frac{0}{0} \), an indeterminate form. Thus, we need to simplify the expression further.
02

Factor the Numerator and Denominator

The numerator \( v^3 - 8 \) is a difference of cubes that can be factored as \( (v-2)(v^2 + 2v + 4) \). The denominator \( v^4 - 16 \) is a difference of squares which can be factored as \( (v^2 - 4)(v^2 + 4) \) and further as \( (v-2)(v+2)(v^2 + 4) \).
03

Simplify the Expression

After factoring, the function becomes \( \frac{(v-2)(v^2 + 2v + 4)}{(v-2)(v+2)(v^2 + 4)} \). We can cancel the \( (v-2) \) terms in the numerator and denominator, noting that \( v eq 2 \) for simplification purposes. This gives us \( \frac{v^2 + 2v + 4}{(v+2)(v^2 + 4)} \).
04

Evaluate the Limit

Now, substitute \( v = 2 \) into the simplified expression: \( \frac{(2)^2 + 2(2) + 4}{(2+2)((2)^2 + 4)} \). This simplifies to \( \frac{4 + 4 + 4}{4(4 + 4)} = \frac{12}{32} = \frac{3}{8} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
When evaluating a limit, like \( \lim _{v \rightarrow 2} \frac{v^{3}-8}{v^{4}-16} \), sometimes substituting the given value directly into the function results in an undefined or indeterminate form, such as \( \frac{0}{0} \). This happens because both the numerator and denominator evaluate to zero. Indeterminate forms are crucial in calculus because they indicate that further analysis is needed to find the actual limit.
  • An indeterminate form implies that the direct substitution method does not work.
  • It is often necessary to simplify the expression further, usually by factoring, to resolve an indeterminate form.
  • Once the problematic parts are simplified or canceled, we can correctly evaluate the limit.
Recognizing when you encounter an indeterminate form is the first step in solving many problems involving limits.
Factoring Polynomials
When faced with an indeterminate form, a common method to resolve it is by factoring polynomials. In the example with the limit \( \lim _{v \rightarrow 2} \frac{v^{3}-8}{v^{4}-16} \), both the numerator and the denominator can be factored to simplify the expression.Factoring is a technique that involves rewriting a polynomial as a product of simpler polynomials. Here are the steps utilized for our specific expressions:
  • The numerator \( v^3 - 8 \) is recognized as a difference of cubes. It can be factored into \( (v - 2)(v^2 + 2v + 4) \).
  • The denominator \( v^4 - 16 \) is a difference of squares. It is factored into \( (v^2 - 4)(v^2 + 4) \). The expression \( v^2 - 4 \) is further factored into \( (v-2)(v+2) \).
By applying these factorization techniques, we rewrite the original complex rational expression, making it easier to simplify and find the limit.
Limit Evaluation
Once the expression is simplified by canceling common factors, usually the next step is evaluating the limit by substitution. After successfully factoring and simplifying \( \frac{(v-2)(v^2 + 2v + 4)}{(v-2)(v+2)(v^2 + 4)} \) to \( \frac{v^2 + 2v + 4}{(v+2)(v^2 + 4)} \), we eliminate the indeterminate form.Here's how we evaluated the limit:
  • Substitute \( v = 2 \) into the simplified expression.
  • Calculate each term step-by-step: \( \frac{2^2 + 2(2) + 4}{(2+2)((2)^2 + 4)} = \frac{12}{4 \times 8} \).
  • This results in \( \frac{12}{32} \), which simplifies to \( \frac{3}{8} \).
Through this careful substitution and simplification, we find the correct value for the limit. This process is fundamental in calculus as it resolves limits that are initially undefined by direct substitution.

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Most popular questions from this chapter

For what value of \(b\) is $$ g(x)=\left\\{\begin{array}{ll}{x,} & {x<-2} \\ {b x^{2},} & {x \geq-2}\end{array}\right. $$ continuous at every \(x ?\)

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