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Practice Problems for 4/10

  1. A citrus farmer usually plants 30 orange trees per acre, and each tree yields 500 oranges when harvested. For each additional tree per acre, each tree produces 10 fewer oranges. How many trees should be planted per acre to maximize the harvest, and what is the yield of this harvest?
    2. Let \(x\) be the number of additional trees planted per acre, and let \(O(x)\) be the number of oranges harvested per acre. Our goal is to maximize \(O(x)\) for \(x\geq 0\).
    \(O(x)=(30+x)(500-10x)\)
    \(O'(x)=200-20x\), so there is one critical point at \(x=10\). \(O''(x)=-20<0\) so by the second derivative test this gives a maximum. The maximum yield is obtained when 40 trees are planted per acre, and the number of oranges harvested is \(O(10)=4,000\).
  2. Which is correct, and WHY?
    1. \( \lim_{x\rightarrow 5}\frac{x-5}{x^3-5}=\lim_{x\rightarrow 5} \frac{1}{3x^2}=\frac{1}{75}\)
    2. \( \lim_{x\rightarrow 5}\frac{x-5}{x^3-5}=\frac{0}{120}=0\)
    Part b is correct.
    L'Hopital's Rule only applies if the limit of both the numerator and denominator are zero, or if both of the limits do not exist because the functions tend to \(\pm \infty\). Thus in part a since the limit of the denominator is not zero, it was not valid to apply L'Hopital's rule.
  3. For each, calculate the limit, or state why it does not exist.
    1. \(\lim_{x\rightarrow \infty} e^{-x}\ln x\)
      2. This limit is indeterminate of the type "\(0\cdot \infty\)." In order to apply L'Hopital's rule we must rewrite the function: \( \lim_{x\rightarrow \infty} e^{-x}\ln x=\lim_{x\rightarrow \infty} \frac{\ln x}{e^x} \). This is an indeterminate limit of the form "\(\frac{\infty}{\infty}\)" so by L'Hopital's rule it is equal to \[ \lim_{x\rightarrow \infty} \frac{1/x}{e^x}=\lim_{x\rightarrow \infty} \frac{1}{xe^x}=0, \] because the numerator tends to the constant 1 and the denominator grows without bound.
    2. \(\lim_{x\rightarrow \infty} x-\sqrt{x^2-1}\)
      3. This limit is indeterminate of the type "\(\infty- \infty\)." In order to apply L'Hopital's rule we must rewrite the function. Multiply the numerator and the denominator by the conjugate to get \( \lim_{x\rightarrow \infty} x-\sqrt{x^2-1}=\lim_{x\rightarrow \infty} \frac{1}{x+\sqrt{x^2-1}}=0.\) (It is actually not necessary to use L'Hopital's rule after rewriting the function.)
    3. \(\lim_{x\rightarrow 0} \frac{e^x-1}{x^2}\)
      4. This limit is indeterminate of the type "\(\frac{0}{0}\)," so we may apply L'Hopital's rule to get \[ \lim_{x\rightarrow 0} \frac{e^x-1}{x^2}=\lim_{x\rightarrow 0} \frac{e^x}{2x} \] Now, the limit of the numerator is 1 and the limit of \(1/2x\) tends to infinity. We therefore conclude that the limit does not exist because the function tends to infinity, much like we would when considering \(\lim_{x\rightarrow 0}\frac{1}{x}\). Note that this conclusion is not due to a limit law.