Math Problem Statement

answer all the question

Evaluate each limit, if it exists for the following problems: a) \(\lim_{{x \to 2}} \frac{{(2 + x)^2 - 16}}{{x - 2}}\) b) \(\lim_{{x \to 6}} \frac{{(3 - x)^2 - 9}}{{x - 6}}\) c) \(\lim_{{x \to 2}} \frac{{49 - (5 + x)^2}}{{x - 2}}\) d) \(\lim_{{x \to 3}} \frac{{\frac{1}{3} - \frac{1}{x}}}{{x - 3}}\) e) \(\lim_{{x \to 2}} \frac{{x^4 - 16}}{{x + 2}}\) f) \(\lim_{{x \to 1}} \frac{{x^2 - 1}}{{x^3 - x^2 - 3x + 3}}\)

Solution

Let's evaluate each limit step-by-step.

a) lim⁡x→2(2+x)2−16x−2\lim_{{x \to 2}} \frac{{(2 + x)^2 - 16}}{{x - 2}}

  1. Simplify the numerator: (2+x)2=4+4x+x2(2 + x)^2 = 4 + 4x + x^2. So, (2+x)2−16=x2+4x−12(2 + x)^2 - 16 = x^2 + 4x - 12.
  2. Factor the expression: (x+6)(x−2)(x + 6)(x - 2).
  3. Cancel out x−2x - 2: (x+6)(x−2)x−2=x+6\frac{{(x + 6)(x - 2)}}{{x - 2}} = x + 6.
  4. Substitute x=2x = 2: 2+6=82 + 6 = 8.

Answer: 88

b) lim⁡x→6(3−x)2−9x−6\lim_{{x \to 6}} \frac{{(3 - x)^2 - 9}}{{x - 6}}

  1. Simplify the numerator: (3−x)2=9−6x+x2(3 - x)^2 = 9 - 6x + x^2, so (3−x)2−9=x2−6x+9(3 - x)^2 - 9 = x^2 - 6x + 9.
  2. Factor the expression: (x−3)(x+3)(x - 3)(x + 3).
  3. **Substitute (x

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Math Problem Analysis

Mathematical Concepts

Calculus
Limits
Factoring
Simplification

Formulas

Difference of Squares: a^2 - b^2 = (a - b)(a + b)
Limit of a Quotient: lim(x→a) (f(x)/g(x)) = lim(x→a) f(x) / lim(x→a) g(x) if lim(x→a) g(x) ≠ 0

Theorems

Limit Laws (Sum, Difference, Product, Quotient Rules)
Squeeze Theorem (if relevant)

Suitable Grade Level

Grades 11-12 (AP Calculus)

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