Math Problem Statement

Evaluate \(\lim_{{x \to 2}} \frac{(x - 3)^{10} - 1}{4 - x^2}\).

Solution

To evaluate the limit

limx2(x3)1014x2\lim_{{x \to 2}} \frac{(x - 3)^{10} - 1}{4 - x^2}

we can proceed with the following steps:

Step 1: Check for Direct Substitution

By substituting x=2x = 2 directly:

  • The numerator becomes (23)101=(1)101=11=0(2 - 3)^{10} - 1 = (-1)^{10} - 1 = 1 - 1 = 0.
  • The denominator becomes 422=44=04 - 2^2 = 4 - 4 = 0.

This results in a 00\frac{0}{0} indeterminate form, so we need to simplify further.

Step 2: Factor the Denominator

Observe that 4x24 - x^2 can be factored as: 4x2=(2x)(2+x)4 - x^2 = (2 - x)(2 + x)

Step 3: Apply the Binomial Expansion for the Numerator

Since (x3)101(x - 3)^{10} - 1 is a difference of two terms, we can consider expanding it. However, this can be quite complex to expand directly. A more efficient approach might be using L'Hôpital's Rule because it is a 00\frac{0}{0} indeterminate form.

Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator with respect to xx:

  1. Differentiate the numerator (x3)101(x - 3)^{10} - 1: ddx((x3)101)=10(x3)9\frac{d}{dx} \left( (x - 3)^{10} - 1 \right) = 10(x - 3)^9

  2. Differentiate the denominator 4x24 - x^2: ddx(4x2)=2x\frac{d}{dx} \left( 4 - x^2 \right) = -2x

Now the limit becomes: limx210(x3)92x\lim_{{x \to 2}} \frac{10(x - 3)^9}{-2x}

Step 5: Substitute x=2x = 2 in the Simplified Expression

Substituting x=2x = 2 into the simplified expression gives: 10(23)922=10(1)94=104=104=52\frac{10(2 - 3)^9}{-2 \cdot 2} = \frac{10(-1)^9}{-4} = \frac{-10}{-4} = \frac{10}{4} = \frac{5}{2}

Final Answer

limx2(x3)1014x2=52\lim_{{x \to 2}} \frac{(x - 3)^{10} - 1}{4 - x^2} = \frac{5}{2}

Would you like further details or have any questions?

Here are five related questions you might find helpful:

  1. How would you solve this limit if the exponent in the numerator was different?
  2. Can L'Hôpital's Rule always be applied in cases of indeterminate forms?
  3. How would the answer change if the exponent were an even number?
  4. What are the conditions necessary to apply L'Hôpital's Rule?
  5. How does the binomial expansion help in solving limit problems?

Tip: When facing an indeterminate form, always check if factoring, expanding, or L'Hôpital’s Rule can simplify the expression.

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Indeterminate Forms

Formulas

L'Hôpital's Rule: \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\) if \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form

Theorems

L'Hôpital's Rule

Suitable Grade Level

Grades 11-12