Calculus Chapter 1.6 Limits Quiz

Limits Evaluation Techniques

• Limit of a Rational Function
  For the function f(x) defined as ( f(x) = \frac{x^2 - 9}{x^2 + 2x - 15} ) at ( x \to 3 ), the limit is computed as ( \frac{3}{4} ).

• Factoring Approach
  Simplifying the expression involves factoring and cancelling common terms:
  ( f(x) = \frac{(x+3)(x-3)}{(x+5)(x-3)} = \frac{x+3}{x+5} ).
  The limit can then be resolved to ( \frac{6}{8} = \frac{3}{4} ).

Limit of Difference Quotients

• Limit Calculation Example
  For the limit of ( \frac{x - 3}{x^3 - 9x} ) as ( x \to 3 ), the result is ( \frac{1}{18} ).

• Rearranging the Expression
  The expression can be factorized:
  ( x^3 - 9x = x(x+3)(x-3) ).
  The limit simplifies to ( \lim_{x \to 3} \frac{1}{x(x+3)} = \frac{1}{1 \cdot 6} = \frac{1}{18} ).

Trigonometric Limits

• Calculating Limit with Sine
  Utilizing known limits, ( \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 ) aids in finding ( \lim_{x \to 0} \frac{\cos(5x)}{8x \cot(2x)} ).
  This limit simplifies to ( \frac{1}{4} ).

• Using Cotangent Form
  The cotangent can be expressed in terms of sine and cosine, allowing manipulation of the limit to achieve the final answer of ( \frac{1}{4} ).

Further Limit Evaluations

• Limit at Negative Values
  For ( \lim_{x \to -4} \frac{x + 4}{x^3 - 16x} ), the limit evaluates to ( \frac{1}{32} ).

• Factoring and Simplifying
  Factoring results in:
  ( x^3 - 16x = x(x - 4)(x + 4) ).
  The limit computation then concludes as ( \frac{1}{1 \cdot (-8)} = \frac{1}{32} ).

Combining Functions

• Tan and Secant Limit
  To find ( \lim_{x \to 0} \frac{\tan(2x)}{6x \sec(3x)} ), the result is ( \frac{1}{3} ).

• Rewriting Using Trigonometric Identities
  By expressing the tangent and secant functions in terms of sine and cosine, the limit simplifies to achieve the value of ( \frac{1}{3} ).

Evaluating Limits for Specific Functions

• Limit at a Point
  For ( f(x) = \frac{x^2 - 4}{x^2 + x - 6} ) at ( x \to 2 ), the limit is ( \frac{4}{5} ).

• Factoring Methodology
  After factoring ( f(x) = \frac{(x-2)(x+2)}{(x-2)(x+3)} ), the limit resolves as:
  ( \frac{2+2}{2+3} = \frac{4}{5} ) as ( x ) approaches 2.