Week 3 Worksheet

  1. Convert 40° to radians. Convert \(\frac{7\pi}{8}\) to degrees.

    Solution

    \(40° \cdot \frac{2\pi}{360°} = \frac{40\pi}{180} = \frac{2\pi}{9}\), meaning 40° is \(\frac{\pi}{9}\) radians.

    \(\frac{7\pi}{8}\cdot \frac{360°}{2\pi} = \frac{7 \cdot 360°}{16} = 157.5°\), so that \(\frac{7\pi}{8}\) radians is 157.5°.

  2. Complete the following table, filling in the sign for each function (if you get stuck, remember you can write everything in terms of sine and cosine!)

    Range \(\sin\theta\) \(\cos\theta\) \(\tan\theta\) \(\sec\theta\) \(\csc\theta\) \(\cot\theta\)
    \(0 < \theta < \pi/2\) \(+\)          
    \(\pi/2 < \theta < \pi\) \(+\)          
    \(\pi < \theta < 3\pi/2\) \(-\)          
    \(3\pi/2 < \theta < 2\pi\) \(-\)          
    Solution
    Range \(\sin\theta\) \(\cos\theta\) \(\tan\theta\) \(\sec\theta\) \(\csc\theta\) \(\cot\theta\)
    \(0 < \theta < \pi/2\) \(+\) \(+\) \(+\) \(+\) \(+\) \(+\)
    \(\pi/2 < \theta < \pi\) \(+\) \(-\) \(-\) \(-\) \(+\) \(-\)
    \(\pi < \theta < 3\pi/2\) \(-\) \(-\) \(+\) \(-\) \(-\) \(+\)
    \(3\pi/2 < \theta < 2\pi\) \(-\) \(+\) \(-\) \(+\) \(-\) \(-\)

  3. Find the period and amplitude of the functions \(f(x) = 10\cos(4x)\) and \(g(x) = -2\sin \left(\frac{\pi}{4}x + 10\right)\).

    Solution
    \(f(x) = 10\cos(4x)\)
    The period is \(\frac{2\pi}{4} = \frac{\pi}{2}\), and the amplitude is 10.
    \(g(x) = -2\sin\left(\frac{\pi}{4}x + 10\right)\)
    The period is \(\frac{2\pi}{\pi/4} = 8\), and the amplitude is 2. (Notice it’s not \(-2\))

  4. For what values of \(\theta\) does \(\sin \theta = -1/2\) and \(\cos \theta = \sqrt{3}/2\)? (I’m asking for all values here, not just in the range \([0, 2\pi)\).)

    Solution

    Looking at the unit circle, the only place this happens in the range \([0, 2\pi)\) is for \(\theta = \frac{11\pi}{6}\). However, we can also add or subtract \(2\pi\) to \(\theta\) any whole number of times, and all the trig functions will have the same value. Thus the possible values can all be written as \(\frac{11\pi}{6} + 2\pi n\), where \(n\) is any integer. (I wouldn’t worry too much if you didn’t write the answer in exactly this format, as long as you expressed this idea in some way)

  5. Compute \(\displaystyle \lim_{x \to 0^-} \frac{|x|}{x}\), \(\displaystyle \lim_{x \to 0^+}\frac{|x|}{x}\) and \(\displaystyle \lim_{x \to 0} \frac{|x|}{x}\).

    Solution

    The key here is to notice that \(|x|\) gives the “positive version” of \(x\). If \(x\) is positive to start with, then \(|x| = x\). If \(x\) is negative, then \(-x\) is positive, and so \(|x| = -x\). The limit from the left only looks at negative values of \(x\), so

    \[ \lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1 \]

    Similarly, the limit from the right only looks at positive values of \(x\):

    \[ \lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1 \]

    Since the limits on either side give different values, the limit \(\lim_{x \to 0} \frac{|x|}{x}\) does not exist.

  6. Let \(g\) be the function

    \[ g(x) = \begin{cases} 5 & \quad x < 0 \\ x^2 - 2 &\quad 0 \leq x \leq 3 \\ 7 &\quad x > 3 \end{cases} \]

    Find \(\displaystyle \lim_{x \to 0} g(x)\) and \(\displaystyle \lim_{x \to 3} g(x)\).

    Solution \begin{align*} \lim_{x \to 0^-} g(x) &= 5 & \lim_{x \to 0^+} g(x) &= 0^2 - 2 = -2 & \lim_{x \to 0} g(x) &\text{ does not exist.} \\ \lim_{x \to 3^-} g(x) &= 3^2 - 2 = 7 & \lim_{x \to 3^+} g(x) &= 7 & \lim_{x \to 3} g(x) &= 7 \end{align*}

  7. Compute \(\displaystyle \lim_{x \to \infty} \frac{9001x^2 + 4x}{x^3 - 1}\).

    Solution

    \(\displaystyle \lim_{x \to \infty} \frac{9001x^2 + 4x}{x^3 - 1} = \lim_{x \to \infty} \frac{9001x^2}{x^3} = \lim_{x \to \infty} \frac{9001}{x} = 0\)

  8. Compute \(\displaystyle \lim_{x \to -\infty} g(x)\) and \(\displaystyle \lim_{x \to \infty} g(x)\), where \(g\) is the following function: \[ g(x) = \begin{cases} \frac{2x + 1}{x - 1} &\quad x < 1 \\ x + 3 &\quad 1 \leq x < 2 \\ \frac{1}{x^2 + 1} &\quad x \geq 2 \end{cases} \]

    Solution

    When we take the limit to \(-\infty\), we only use the left-most “piece” of \(g\). When we take the limit to \(\infty\), we only use the right-most “piece.” \[ \lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} \frac{2x + 1}{x - 1} = \lim_{x \to -\infty} \frac{2x}{x} = 2 \] \[ \lim_{x \to \infty} g(x) = \lim_{x \to \infty} \frac{1}{x^2 + 1} = 0 \]

  9. Compute \(\displaystyle \lim_{x \to \infty} \frac{\sin x}{x}\)

    Solution

    There’s no algebraic trick here. Just notice that while the denominator goes to \(\infty\), the numerator stays relatively small, since \(-1 \leq \sin x \leq 1\) for all \(x\). Thus the limit is 0.

    More precisely, this ineqaulity tells us

    \[ \frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \]

    But as \(x \to \infty\), both \(-1/x\) and \(1/x\) go to 0. Thus \(\frac{\sin x}{x}\) must also go to 0, since it is “squeezed” in between the other two functions. (visualization)

  10. Does there exist a number \(k\) such that \(\displaystyle \lim_{x \to 2} \frac{3x^2 + kx - 2}{x^2 - 3x + 2}\) exists? Find such a \(k\) or explain why it can’t exist. Then do the same for \(\displaystyle \lim_{x \to 2}\frac{x^2 + kx}{x^2 - 4x + 4}\).

    Solution

    The denominator of \(\frac{3x^2 + kx - 2}{x^2 - 3x + 2}\) factors as \(x^2 - 3x + 2 = (x - 2)(x - 1)\). If we want this limit to exist as \(x \to 2\), our only hope is that the numerator also factors, where one of the factors is \(x - 2\), so that this function has a hole at \(x = 2\). If not, then it will have a vertical asymptote, and the limit won’t exist. For \(x - 2\) to be a factor of the numerator, we need the numerator to be 0 when we plug in 2. That means

    \begin{align*} 3(2)^2 + k(2) - 2 &= 0 \\ k &= -5 \end{align*}

    So we should try \(k = -5\):

    \[ \lim_{x \to 2} \frac{3x^2 - 5x - 2}{x^2 - 3x + 2} = \lim_{x \to 2} \frac{(3x + 1)(x - 2)}{(x - 1)(x - 2)} = \lim_{x \to 2} \frac{3x + 1}{x - 1} = \frac{6 + 1}{2 - 1} = 7 \]

    So the limit exists for \(k = -5\).

    For the second limit, the denominator factors as \(x^2 -4x + 4 = (x - 2)(x - 2)\). In order for the limit to exist, the numerator needs to have two factors of \(x - 2\), to cancel out both of these. Again the numerator needs to be 0 when \(x = 2\), so if any \(k\) is going to work, it better satisfy the equation

    \begin{align*} (2)^2 + k(2) &= 0 \\ k &= -2 \end{align*}

    So the only possible thing that \emph{could} work is \(k = -2\). If we try this, we get

    \[ \lim_{x \to 2} \frac{x^2 - 2x}{x^2 - 4x + 4} = \lim_{x \to 2} \frac{x(x - 2)}{(x - 2)(x - 2)} = \lim_{x \to 2} \frac{x}{x - 2} \]

    But now the denominator \(x - 2\) goes to 0 at \(x = 2\), but the numerator doesn’t, so we have a vertical asymptote at \(x = 2\), and so the limit doesn’t exist. Since \(k = -2\) was the only thing that could work, and it didn’t work, there can’t be any value of \(k\) that makes the limit exist.

  11. In class, we learned \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\). This challenge exercise will outline how to show this using elementary geometry and algebra.

    1. Use geometry to argue that for small positive values of \(x\),

      \[ \sin x \leq x \leq \tan x. \]

      (How can we compare an angle to a length? Remember that radians measure arc length around the unit circle!)

    2. Manipulate these inequalities using algebra to get

      \[ \cos x \leq \frac{\sin x}{x} \leq 1 \]

    3. Compute the limit of \(\cos x\) as \(x \to 0\). Give an intuitive argument why this means \(\frac{\sin x}{x}\) must converge to \(1\) as \(x \to 0^+\). (We only get from the right, since in the first step we assumed \(x\) was positive to get the geometry right).
    4. Adapt this argument for negative values of \(x\) to get the two-sided limit.
    Solution

    Come to my office hours for the full solution :). There is a nice clear proof published online.

    A note of caution: the boxed proof at the start is a sketchy and/or bad proof, and the next few paragraphs are the author of this page critiquing it. The clear and simple proof starts with “Personally, I find the proof of (*)…” The proof ends with the picture of the bow.