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JA2. Limits

Question 1

1. Use the limit laws to solve the problem: \(\displaystyle \lim_{x \to 3} \frac{x^4-x^2+ 3}{x^2 +2}\)

\[ \begin{aligned} \lim_{x \to 3} \frac{x^4-x^2+ 3}{x^2 +2} &= \lim_{x \to 3} \frac{x^2(x^2-1)+ 3}{x^2 +2} \\ &= \lim_{x \to 3} \frac{9(9-1) + 3}{9 +2} \\ &= \lim_{x \to 3} \frac{72 + 3}{11} \\ &= \lim_{x \to 3} \frac{75}{11} \\ \end{aligned} \]

Question 2

2. Explain the continuity of a function at any point. Explain the procedure to check continuity using a simple example.

For a function to be continuous at a point, it must (Strang & Herman, n.d.):

  • Be defined at that point,
  • Its limit must exist at the point,
  • The value of the function at that point must equal the value of the limit at that point.

As a simple example, I will use the function \(f(x) = x^2 +1\) and check its continuity at \(x=0\):

  • The function is defined at \(x=0\) as its domain is \(\mathbb{R}\).
  • The limit of the function at \(x=0\) is \(\lim_{x \to 0} f(x) = \lim_{x \to 0} x^2 +1 = 1\).
  • The value of the function at \(x=0\) is \(f(0) = 0^2 +1 = 1\).

As the limit of the function at \(x=0\) equals the value of the function at \(x=0\), the function is continuous at this point.

As a counter example, I will use the function \(f(x) = \frac{1}{x}\) and check its continuity at \(x=0\):

  • The function is not defined at \(x=0\) as its domain is \(\mathbb{R} \setminus \{0\}\).
  • The limit of the function at \(x=0\) is \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x} = \infty\).

The function is not continuous at \(x=0\) as the limit of the function at \(x=0\) does not exist and the function is not defined at this point.

Question 3

3. A rock is dropped from a height of 16 ft. It is determined that its height (in feet) above ground t seconds later \((for 0≤t≤3)\) is given by \(s(t)=-2t^2 + 16\). Find the average velocity of the rock over [0.2,0.21] time interval.

At \(t=0.2\):

  • The rock was at \(s(0.2) = -2(0.2)^2 + 16 = 15.92\) feet above the ground.
  • The rock has traveled \(s(0.2) - s(0) = 15.92 - 16 = -0.08\) feet.
  • The average velocity of the rock over the time interval \(V_{0 → 0.2} = \frac{s(0.2) - s(0)}{0.2 - 0} = \frac{-0.08}{0.2} = -0.4\) feet per second.

At \(t=0.21\):

  • The rock was at \(s(0.21) = -2(0.21)^2 + 16 = 15.8918\) feet above the ground.
  • The rock has traveled \(s(0.21) - s(0) = 15.8918 - 16 = -0.1082\) feet.
  • The average velocity of the rock over the time interval \(V_{0 → 0.21} = \frac{s(0.21) - s(0)}{0.21 - 0} = \frac{-0.1082}{0.21} = -0.5152\) feet per second.

But we are only interested in velocity over the time interval \([0.2,0.21]\), so the average velocity of the rock over this time interval is

\[ \begin{aligned} V_{0.2 → 0.21} &= \frac{s(0.21) - s(0.2)}{0.21 - 0.2} \\ &= \frac{-0.5152 - (-0.4)}{0.21 - 0.2} \\ &= -1.152 \\ \end{aligned} \]

feet per second.

Question 4

4. Evaluate each of the following limits. Identify any vertical asymptotes of the function \(f(x) = \frac{1}{(x-4)^2 (x+3)}\).

(i).

\[ \begin{aligned} \displaystyle \lim_{x \to 4^+} f(x) &= \lim_{x \to 4^+} \frac{1}{(x-4)^2 (x+3)} \\ &= + \infty \end{aligned} \]

(ii).

\[ \begin{aligned} \displaystyle \lim_{x \to 4^-} f(x) &= \lim_{x \to 4^-} \frac{1}{(x-4)^2 (x+3)} \\ &= + \infty \end{aligned} \]

(iii).

\[ \begin{aligned} \displaystyle \lim_{x \to 4} f(x) &= \lim_{x \to 4} \frac{1}{(x-4)^2 (x+3)} \\ &= + \infty \end{aligned} \]

The square of the nominal \((x-4)^2\) in the denominator will always be positive, thus, the function approaches positive infinity as \(x\) approaches \(4\) from both sides. The function has a vertical asymptote at \(x=4\) and \(x=-3\).

Question 5

5. Evaluate the trigonometric limit \(\displaystyle \lim_{ \theta \to 0} \frac{2sin^2 \theta }{1 - cos \theta }\).

\[ \begin{aligned} \lim_{ \theta \to 0} \frac{2sin^2 \theta }{1 - cos \theta } &= \lim_{ \theta \to 0} \frac{2sin^2 \theta }{1 - cos \theta } \times \frac{1 + cos \theta }{1 + cos \theta } \\ &= \lim_{ \theta \to 0} \frac{2sin^2 \theta (1 + cos \theta )}{1 - cos^2 \theta } \\ &= \lim_{ \theta \to 0} \frac{2sin^2 \theta (1 + cos \theta )}{sin^2 \theta } \\ &= \lim_{ \theta \to 0} 2(1 + cos \theta ) \\ &= 2(1 + cos 0) \\ &= 2(1 + 1) \\ &= 4 \\ \end{aligned} \]

Question 6

6. Find the value of k that makes the following function is continuous over the given interval. \(f(x) = {5x + 4, x \leq k \brace 3x - 4, k < x < 7}\).

  • The \(K\) is the point where the function changes from \(5x + 4\) to \(3x - 4\).
  • The \(x=k\) belongs to the first branch, but as \(x\) approaches \(k\) from the right, it belongs to the second branch.
  • Thus, the function is continuous at \(k\) when \(5x + 4 = 3x - 4\) for \(x=k\).
  • We can write this as \(5k + 4 = 3k - 4\) and solve for \(k\) to get \(k = -4\).
  • Let’s prove that the function is continuous at \(x=-4\):
    • \(\displaystyle \lim_{x \to -4^+} f(x) = \lim_{x \to -4^+} 3x - 4 = -16\)
    • \(f(-4) = 5(-4) + 4 = -16\)
    • The function is obviously defined at \(x=-4\).
  • Thus, we have proven that the function is continuous at \(k=-4\).

Question 7

7. Determine at the point 5, if the following function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other. \(f(x) = \frac{\left| \begin{matrix} x - 5 \end{matrix} \right| }{x - 5 }\).

  • The above function is undefined at \(x=5\) as the denominator is zero at this point, so we can conclude that it cannot be continuous at this point.
  • Let’s evaluate the limit of the function at \(x=5\) (from both sides):
    • \(\displaystyle \lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} \frac{\left| \begin{matrix} x - 5 \end{matrix} \right| }{x - 5 } = \lim_{x \to 5^+} \frac{x - 5 }{x - 5 } = 1\)
    • \(\displaystyle \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} \frac{\left| \begin{matrix} x - 5 \end{matrix} \right| }{x - 5 } = \lim_{x \to 5^-} \frac{-(x - 5) }{x - 5 } = -1\)
  • As the right limit does not equal the left limit, the function is discontinuous at \(x=5\) and the discontinuity is a jump discontinuity.

References