BC Calculus: Extra Limits Skills Practice. §2.2 and §2.3

Evaluate the following limits:

  1. \(\displaystyle\lim_{x \to 2}\frac{x^2 - 4}{x^2 + 4} = \)
    2. Solution = 0. Substitute in and evaluate to get 0/8
  2. \(\displaystyle\lim_{x \to \infty }\frac{4 - x^2}{x^2 - 1} =\)
    3. Solution = -1 (Ratio of coefficients of dominant terms.)
  3. \(\displaystyle\lim_{x \to 3}\frac{x - 3}{x^2 - 2x - 3} =\)
    4. Solution = \(\frac{1}{4}\) (Factor denominator, reduce, substitue.)
  4. \(\displaystyle\lim_{x \to 0 }\frac{|x|}{x} = \)
    5. Solution D. N. E. (This is \(y = 1\) to the right of the y-axis, and \(y = -1\) to the left of the y-axis. Left- and right-handed limits disagree.)
  5. \(\displaystyle\lim_{x \to 0 } \frac {\sin{(3x)}}{\sin{(4x)}} = \)
    6. Solution = \(\frac{3}{4}\). (Using \(\displaystyle\lim_{x \to 0} \frac{\sin{x}}{x} = 1\) and substitution to get \( \displaystyle \lim_{x \to 0} \frac{\sin{(ax)}}{bx} = \frac{a}{b}\).)
  6. \(\displaystyle\lim_{x \to \infty } \sin {\left( \frac{1}{x} \right) } = \)
    7. Solution = 0. (Note this is \(x \to \infty\), not 0.)
  7. \(\displaystyle\lim_{x \to 0 }\sin{\left( \frac{1}{x} \right) } = \)
    8. Solution D. N. E. (Oscillates infinitely.)
  8. \(\displaystyle\lim_{x \to 25}\frac{\sqrt{x} - 5}{x - 25} = \)
    9. Solution = \(\frac{1}{\sqrt{25} + 5}\). (Rationalize numerator -- or factor denominator -- then reduce. No need to simplify the denominator in the final answer.)
  9. \(\displaystyle\lim_{x \to \infty }\frac{3x^2 + 27}{x^3 - 27} = \)
    10. Solution = 0. (Dominant term is in denominator.)
  10. \(\displaystyle\lim_{x \to 0} x \csc{x} = \)
    11. Solution = 1. (This is \(\frac{x}{\sin{x}}\), which just approaches the reciprocal of 1.)
  11. \(\displaystyle\lim_{x \to 0 }|x| = \)
    12. Solution = 0. (An uninteresting question, until you get to the next chapter.)