Problem 31 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{x \rightarrow-1} \frac{x^{3}-x^{2}-5 x-3}{x^{4}+2 x^{3}-x^{2}-4 x-2}$$

Step-by-Step Solution

Answer

Question: Evaluate the limit of the given rational function as x approaches -1: $$\lim _{x \rightarrow-1} \frac{x^{3}-x^{2}-5 x-3}{x^{4}+2 x^{3}-x^{2}-4x-2}$$ Answer: 0

Step 1: Check for continuity

Check if the given function is continuous at x = -1. If the function is continuous, then the limit can be directly evaluated. If the degree of the polynomial in the numerator is less than the denominator, the limit will be 0. If the degrees are equal, the limit will be the ratio of the leading coefficients. If the degree of the polynomial in numerator is greater than the denominator, the limit will be infinity or undefined. In our case, the degree of the denominator is greater than the degree of the numerator. However, let's plug x = -1 into the function to check if any simplifications can be made.

Step 2: Substitute x = -1 into the function

Now substitute x = -1 into the function and simplify the numerator and the denominator: $$\frac{(-1)^{3}-(-1)^{2}-5(-1)-3}{(-1)^{4}+2(-1)^{3}-(-1)^{2}-4(-1)-2}$$ By simplifying, we get: $$\frac{-1 - 1 + 5 -3}{1 - 2 - 1 + 4 -2}$$ The result is still an indeterminate form (0/0). We need to continue factoring the numerator and denominator to simplify the expression further.

Step 3: Factor out common terms

We can try factoring the numerator and denominator to find a common factor that would cancel out. Upon inspection, we see that both the numerator and denominator have a common factor of (x+1). The numerator can be factored as: $$x^{3}-x^{2}-5 x-3 = (x+1)(x^{2}-2x-3)$$ The denominator can be factored as: $$x^{4}+2 x^{3}-x^{2}-4x-2 = (x+1)(x^{3}+x^{2}-2x-2)$$ Plugging these factors back into our limit expression: $$\lim _{x \rightarrow-1} \frac{(x+1)(x^{2}-2x-3)}{(x+1)(x^{3}+x^{2}-2x-2)}$$

Step 4: Cancel out (x+1) terms

Since we found a common factor of (x+1) in both numerator and denominator, we can cancel it out: $$\lim _{x \rightarrow-1} \frac{x^{2}-2x-3}{x^{3}+x^{2}-2x-2}$$ Now the function is continuous at x=-1. We can substitute x = -1 into the simplified function to find the limit.

Step 5: Substitute x = -1 and find the limit

Substitute x = -1 into the simplified function: $$\frac{(-1)^{2}-2(-1)-3}{(-1)^{3}+(-1)^{2}-2(-1)-2}$$ Simplify the expression: $$\frac{1 + 2 -3}{-1 +1 + 2 -2}=0$$ So the limit of the given function as x approaches -1 is: $$\lim _{x \rightarrow-1} \frac{x^{3}-x^{2}-5 x-3}{x^{4}+2 x^{3}-x^{2}-4x-2}=0$$

Key Concepts

Continuity

Continuity is a key concept when it comes to evaluating limits. A function is said to be continuous at a point if the limit of the function as it approaches the point equals the function's value at that point.
For example, if we want to find the limit of a function as it approaches -1, we first check if the function is continuous at x = -1.
A continuous function does not have any jumps, holes, or asymptotic behavior at that point.

• If the function is continuous at x = -1, we can directly substitute x = -1 into the function to get the limit.
• If the function is not continuous, we need to find an alternative, like factoring or using L'Hospital's Rule, if appropriate, to solve the limit.

In our exercise, after simplifying, the function becomes continuous at x = -1, allowing us to directly find the limit.

Factorization

Factorization plays an important role in simplifying complex expressions, especially when handling polynomials.
The goal is to rewrite the polynomial in a simpler form, such as a product of factors, which can often resolve indeterminate forms in limits.
For instance, in the exercise given, we have two polynomials, one in the numerator and one in the denominator.

• By factoring these polynomials, we can identify and cancel any common factors that were causing the indeterminate form 0/0.
• In the example, both the numerator and denominator share a common factor of \((x+1)\), which when canceled, simplifies the expression significantly.

This simplification often transforms an undefined expression into a form where the limits can be readily evaluated.

Polynomial Degrees

Understanding polynomial degrees is crucial for evaluating limits, especially when comparing polynomials in numerators and denominators.
The degree of a polynomial is the highest power of the variable present in the expression.

• If the polynomial in the numerator has a lower degree than the denominator, the limit tends toward zero.
• If the polynomial degrees are equal, the limit is the ratio of their leading coefficients.
• If the numerator's degree is higher, the limit becomes infinity or is undefined.

In our exercise, the denominator has a higher degree than the numerator, hinting the limit might be zero after simplification.
This understanding helps in predicting the outcome of the limit evaluation, guiding the steps needed to simplify and compute the limit.

Indeterminate Forms

Indeterminate forms often arise when substituting a point into a function, yielding an expression like 0/0, among other forms. Such forms do not have a defined limit without further work.

• In calculus, these require simplification or alternative approaches such as factoring, L'Hospital's Rule, or algebraic manipulation.
• In the provided exercise, substituting x = -1 initially gives the indeterminate form 0/0, indicating that simple substitution isn't enough and more work, like factoring, is needed.

Once the indeterminate factors are simplified or canceled out, the expression becomes determined, and the limit can be found.