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Practice Problems for 2/4

  1. The height of a ball thrown into the air with velocity of 40 ft/s is given by \(h(t)=40t-16t^2\). Find the average velocity for the give time intervals:
    1. \([.5,1]\)

      2. \(\frac{h(1)-h(.5)}{1-.5}=\frac{24-16}{.5}=16\, ft/s\).
    2. \([1,2]\)

      3. \( \frac{h(2)-h(1)}{2-1}=16-24=-8\, ft/s\)
    3. Interpret the sign (positive or negative) of each of your answers.
      On [.5,1] the ball is moving upward, increasing in height. In [1,2] the ball starts falling and is on average moving down toward the ground.

  2. If \(f(x)=\frac{x^2-4}{x-2}\) and \(g(x)=x+2\),
    1. True or False: we can say that the functions \(f\) and \(g\) are equal. (Write one sentence justifying your response.)

      2. False- \(x=2\) is not in the domain of \(f\), but it is in the domain of \(g \). The functions cannot be equal since they have different domains.
    2. True or False: we can say that \(\lim_{x\rightarrow c}f(x)= \lim_{x\rightarrow c}g(x)\) for every real number \(c\). (Write one sentence justifying your response.)

      3. True- Since \(f(x)=g(x)\) except at \(x=2\), it is clear that \(\lim_{x\rightarrow c}f(x)= \lim_{x\rightarrow c}g(x)\), for \(x \neq 2\). Furthermore, since the functions agree on every point containing \(x=c\), except for at \(c\) itself, it follows that \(\lim_{x\rightarrow c}f(x)= \lim_{x\rightarrow c}g(x)\) for \(x=2\) as well.

  3. Find the value of the limit, or explain why the limit does not exist.
    4. $$ \begin{align*} \lim_{x\rightarrow 1}\frac{x^2+x+1}{x^3-1}&= \lim_{x\rightarrow 1}\frac{x^2+x+1}{(x-1)(x^2+x+1)}\\ &=\lim_{x\rightarrow 1}\frac{1}{x-1}. \end{align*}$$ Thus, the limit does not exist as \(\frac{1}{x-1}\) tends to infinity for \(x>1\) and toward negative infinity for \(x < 1 \).