Question

\(\operatorname{Lt}_{x \rightarrow-1} \frac{x^{5}+1}{x+1}=\) (1) 1 (2) \(-5\) (3) 5 (4) None of these

Short answer

Answer: (1) 1

Step-by-step solution

Identify the Limit

We need to find the limit of the function as x approaches -1. The function is given by: \(\frac{x^5 + 1}{x + 1}\). So, find the limit as x approaches -1: $$ \lim_{x \to -1} \frac{x^5 + 1}{x + 1} $$

Check Direct Substitution

Before using any algebraic manipulation, It's always good to try and plug the value of x directly into the function to see if it is continuous at the point. If it is, then we are done. Let's plug \(x=-1\): $$ \frac{(-1)^5 + 1}{(-1) + 1} = \frac{-1+1}{0} $$ The denominator becomes 0, while the numerator doesn't. It means direct substitution was not possible. We have to try a different method.

Factoring the Function

We factor the numerator using a standard technique. Let \(a = x^5+1 = (x^5-(-1))\). Observe that this expression looks like a difference of squares, but raised to an odd power instead. To factor it properly, we can use a standard formula from algebra: $$ a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+ ab^{n-2} + b^{n-1}), \ \text{when n is odd} $$ Applying this formula will make us consider \(a=x\) and \(b=1\), with \(n=5\). Therefore, the function becomes: $$ \frac{x^5 + 1}{x + 1} = \frac{(x-(-1))(x^4 + x^3 + x^2 + x + 1)}{x+1} $$

Simplify and Calculate the Limit

Since we've factored the function correctly, we can cancel out \((x+1)\) from both numerator and denominator. $$ \lim_{x \to -1} \frac{x^5 + 1}{x + 1} = \lim_{x \to -1} (x^4 + x^3 + x^2 + x + 1) $$ Now, we can directly substitute our limit value \(x=-1\): $$ \lim_{x \to -1} (x^4 + x^3 + x^2 + x + 1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 $$

Finalize the Answer

After substituting the values, calculate the limit: $$ (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 $$ So, the correct answer is \(\boxed{\text{(1) } 1}\).

Key concepts

Rational Functions

Rational functions are expressions that can be written as a fraction where both the numerator and the denominator are polynomials. In any rational function, care must be taken to ensure the denominator is never zero.
Understanding rational functions helps with solving limits because they often appear when working with these types of problems.

• General Form: \( \frac{P(x)}{Q(x)} \)
• The domain is all real numbers except values that make \( Q(x) = 0 \).
• As \( x \) approaches points that make \( Q(x) = 0 \), the limit may not exist, unless numerator cancels appropriately.

In this particular problem, the rational function \( \frac{x^5 + 1}{x + 1} \) needs evaluation as \( x \) approaches \(-1\). Direct substitution shows the denominator becomes zero. This signals the need for polynomial division or algebraic manipulation to handle the limit.

Polynomial Division

Polynomial division is a powerful tool when dealing with limits, especially when the direct substitution results in an indeterminate form like \( \frac{0}{0} \).
When a polynomial in the numerator can be divided by a polynomial in the denominator, factoring or division can simplify the rational function.

• Long Division: Used when polynomials don't share obvious factors.
• Synthetic Division: A shortcut when the divisor is linear and has the form \( x-c \).
• Factoring: Divides the polynomial expression into products of its factors.

In this exercise, simplifying \( \frac{x^5 + 1}{x + 1} \) involved factoring, since the expression in the numerator matches a common algebraic identity.This creates an opportunity to cancel out the \( x + 1 \) term in both the numerator and denominator. The simplification then permits direct evaluation.

Algebraic Manipulation

Successfully solving limits involving rational functions often requires skillful algebraic manipulation. This involves rearranging or transforming the function to expose a clearly evaluable limit.
One valuable technique for this task is factoring the polynomial. When one polynomial divides evenly into another—as with our numerator \( x^5 + 1 \) —factoring can reveal hidden simplifications.

• Identify common algebraic identities, like the sum of cubes or squares.
• Apply formula \( a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+ ab^{n-2} + b^{n-1}) \) where \( n \) is odd, to simplify.
• Focus on canceling common factors between numerator and denominator.

This exercise uses such a formula to reduce \( x^5 + 1 \) into a product of lower-degree polynomials. This simplification transforms an indeterminate limit into a polynomial directly calculable through substitution.