Math Problem Statement

Evaluate the limit: \lim_{x \to 5} \frac{(6x - 6)^{-1} - 24^{-1}}{x - 5}

Solution

To evaluate the limit

limx5(6x6)1241x5,\lim_{x \to 5} \frac{(6x - 6)^{-1} - 24^{-1}}{x - 5},

we can recognize that this expression is in the form of a limit that resembles the definition of the derivative of a function at a point. However, let's simplify the expression step-by-step using algebraic manipulation and limit properties.

Step 1: Simplify the Terms

Rewrite the given limit:

(6x6)1=16x6and241=124.(6x - 6)^{-1} = \frac{1}{6x - 6} \quad \text{and} \quad 24^{-1} = \frac{1}{24}.

So, the expression becomes:

16x6124x5.\frac{\frac{1}{6x - 6} - \frac{1}{24}}{x - 5}.

Step 2: Find a Common Denominator

Combine the fractions in the numerator:

16x6124=24(6x6)(6x6)24=246x+6(6x6)24.\frac{1}{6x - 6} - \frac{1}{24} = \frac{24 - (6x - 6)}{(6x - 6) \cdot 24} = \frac{24 - 6x + 6}{(6x - 6) \cdot 24}.

Simplify the numerator:

246x+6=306x.24 - 6x + 6 = 30 - 6x.

So the expression now becomes:

306x(6x6)24x5.\frac{\frac{30 - 6x}{(6x - 6) \cdot 24}}{x - 5}.

Step 3: Simplify Further

Rewrite the expression as:

306x(6x6)24(x5).\frac{30 - 6x}{(6x - 6) \cdot 24 \cdot (x - 5)}.

Factor out 6-6 from 306x30 - 6x:

306x=6(x5).30 - 6x = -6(x - 5).

Substituting this back into the expression:

6(x5)(6x6)24(x5).\frac{-6(x - 5)}{(6x - 6) \cdot 24 \cdot (x - 5)}.

Cancel the (x5)(x - 5) terms:

6(6x6)24.\frac{-6}{(6x - 6) \cdot 24}.

Step 4: Evaluate the Limit as x5x \to 5

Substitute x=5x = 5:

6x6=6(5)6=24.6x - 6 = 6(5) - 6 = 24.

So the limit simplifies to:

62424=6576=196.\frac{-6}{24 \cdot 24} = \frac{-6}{576} = -\frac{1}{96}.

Final Answer:

limx5(6x6)1241x5=196.\lim_{x \to 5} \frac{(6x - 6)^{-1} - 24^{-1}}{x - 5} = -\frac{1}{96}.

Do you have any questions, or would you like further clarification on any steps? Here are some related questions that might help you:

  1. How can I apply the definition of the derivative using limits?
  2. What are other forms of limits that resemble the derivative definition?
  3. How do you deal with indeterminate forms in limits?
  4. What techniques can be used to simplify complex fractions?
  5. How can factoring help in evaluating limits?

Tip: When you encounter complex fractions in limits, try to simplify by combining fractions and factoring terms—it often makes evaluating the limit easier!

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Math Problem Analysis

Mathematical Concepts

Limits
Algebra
Derivatives

Formulas

Limit definition of derivative
Algebraic manipulation for simplifying fractions

Theorems

Limit properties
Derivative definition

Suitable Grade Level

Grades 11-12