Math Problem Statement

Evaluate the following limit: \[ \lim_{x \to 1} \frac{x^9 - 1}{x^2 - 1} \] Enter -I if your answer is -∞, enter I if your answer is ∞, and enter DNE if the limit does not exist.

Solution

To evaluate the limit

limx1x91x21,\lim_{x \to 1} \frac{x^9 - 1}{x^2 - 1},

we can proceed as follows:

Step 1: Factorize the Expressions

Both the numerator and the denominator can be factorized as differences of powers.

  1. Numerator: x91x^9 - 1 is a difference of powers and can be factored as: x91=(x1)(x8+x7+x6+x5+x4+x3+x2+x+1).x^9 - 1 = (x - 1)(x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1).

  2. Denominator: x21x^2 - 1 is a difference of squares and can be factored as: x21=(x1)(x+1).x^2 - 1 = (x - 1)(x + 1).

Step 2: Simplify the Expression

Now we substitute the factored forms into the expression:

x91x21=(x1)(x8+x7+x6+x5+x4+x3+x2+x+1)(x1)(x+1).\frac{x^9 - 1}{x^2 - 1} = \frac{(x - 1)(x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)}{(x - 1)(x + 1)}.

We can cancel out (x1)(x - 1) from both the numerator and the denominator (assuming x1x \neq 1):

=x8+x7+x6+x5+x4+x3+x2+x+1x+1.= \frac{x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1}{x + 1}.

Step 3: Substitute x=1x = 1 into the Simplified Expression

Now, we substitute x=1x = 1 in the remaining expression:

=18+17+16+15+14+13+12+1+11+1.= \frac{1^8 + 1^7 + 1^6 + 1^5 + 1^4 + 1^3 + 1^2 + 1 + 1}{1 + 1}.

Each power of 1 is simply 1, so we get:

=1+1+1+1+1+1+1+1+12=92.= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1}{2} = \frac{9}{2}.

Final Answer

The limit is:

limx1x91x21=92.\lim_{x \to 1} \frac{x^9 - 1}{x^2 - 1} = \frac{9}{2}.

Would you like more details on any step or have further questions?


Here are some related questions to expand your understanding:

  1. How do you factor expressions involving higher powers, like xn1x^n - 1?
  2. What are other methods to evaluate limits involving indeterminate forms?
  3. How does L'Hôpital's Rule apply to limits with 00\frac{0}{0} forms?
  4. What are some other examples of using the difference of squares in limit problems?
  5. How can partial fractions be useful in evaluating limits?

Tip: When evaluating limits that result in 00\frac{0}{0}, try factoring the numerator and denominator first to simplify the expression before applying other techniques.

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Math Problem Analysis

Mathematical Concepts

Limits
Factoring
Difference of Powers
Difference of Squares

Formulas

Difference of powers: x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ... + 1)
Difference of squares: x^2 - 1 = (x - 1)(x + 1)

Theorems

Limit Laws
Factoring Techniques

Suitable Grade Level

Grades 11-12