Coordinated Calculus

a. We first construct a graph of \(f\) along with tables of values near \(c = -1\) and \(c = -2\text{.}\)

\(x\)	\(f(x)\)
\(-0.9\)	\(2.9\)
\(-0.99\)	\(2.99\)
\(-0.999\)	\(2.999\)
\(-0.9999\)	\(2.9999\)
\(-1.1\)	\(3.1\)
\(-1.01\)	\(3.01\)
\(-1.001\)	\(3.001\)
\(-1.0001\)	\(3.0001\)

\(x\)	\(f(x)\)
\(-1.9\)	\(3.9\)
\(-1.99\)	\(3.99\)
\(-1.999\)	\(3.999\)
\(-1.9999\)	\(3.9999\)
\(-2.1\)	\(4.1\)
\(-2.01\)	\(4.01\)
\(-2.001\)	\(4.001\)
\(-2.0001\)	\(4.0001\)

From Table1.22, it appears that we can make \(f\) as close as we want to 3 by taking \(x\) sufficiently close to \(-1\text{,}\) which suggests that \(\lim_{x \to -1} f(x) = 3\text{.}\) This is also consistent with the graph of \(y=f(x)\) seen in Figure1.24. To see this a bit more rigorously and from an algebraic point of view, consider the formula for \(f\text{:}\) \(f(x) = \frac{4-x^2}{x+2}\text{.}\) As \(x\) approaches \(-1\text{,}\) the numerator, \(4-x^2\text{,}\) of \(f\) approaches \(4 - (-1)^2 = 3\text{,}\) and the denominator, \(x+2\text{,}\) of \(f\) approaches \(-1 + 2 = 1\text{.}\) Hence \(\lim_{x \to -1} f(x) = \frac{3}{1} = 3\text{.}\)

The situation is more complicated when \(x\) approaches \(-2\) because \(f(-2)\) is not defined. If we try to use a similar algebraic argument regarding the numerator and denominator, we observe that as \(x\) approaches \(-2\text{,}\) the numerator \((4-x^2)\) approaches \((4 - (-2)^2) = 0\text{,}\) and the denominator \((x+2)\) approaches \((-2 + 2) = 0\text{,}\) so as \(x\) approaches \(-2\text{,}\) the numerator and denominator of \(f\) both tend to 0. We call \(\frac{0}{0}\) an indeterminate form. This tells us that there is somehow more work to do. From Table1.23 and Figure1.24, it appears that \(f\) should have a limit of \(4\) at \(x = -2\text{.}\)

To see algebraically why this is the case, observe that

It is important to observe that because we are taking the limit as \(x\) approaches \(-2\text{,}\) we are considering \(x\) values that are close, but not equal, to \(-2\text{.}\) Since we never actually allow \(x\) to equal \(-2\text{,}\) the quotient \(\frac{2+x}{x+2}\) has value 1 for every possible value of \(x\text{.}\) Thus, we can simplify the most recent expression above, and find that

This limit is now easy to determine, and its value is \(4\text{.}\) Thus, from several points of view we've seen that \(\lim_{x \to -2} f(x) = 4\text{.}\)

b. Next we turn to the function \(g\text{,}\) and construct two tables and a graph.

\(x\)	\(g(x)\)
\(2.9\)	\(0.84864\)
\(2.99\)	\(0.86428\)
\(2.999\)	\(0.86585\)
\(2.9999\)	\(0.86601\)
\(3.1\)	\(0.88351\)
\(3.01\)	\(0.86777\)
\(3.001\)	\(0.86620\)
\(3.0001\)	\(0.86604\)

\(x\)	\(g(x)\)
\(-0.1\)	\(0\)
\(-0.01\)	\(0\)
\(-0.001\)	\(0\)
\(-0.0001\)	\(0\)
\(0.1\)	\(0\)
\(0.01\)	\(0\)
\(0.001\)	\(0\)
\(0.0001\)	\(0\)

First, as \(x\) approaches \(3\text{,}\) it appears from the values in Table1.25 that the function is approaching a number between \(0.86601\) and \(0.86604\text{.}\) From the graph in Figure1.27 it appears that \(g(x)\) approaches \(g(3)\) as \(x\) approaches \(3\text{.}\) The exact value of \(g(3) = \sin(\frac{\pi}{3})\) is \(\frac{\sqrt{3}}{2}\text{,}\) which is approximately 0.8660254038. This is convincing evidence that

As \(x\) approaches \(0\text{,}\) we observe that \(\frac{\pi}{x}\) does not behave in an elementary way. When \(x\) is positive and approaching zero, we are dividing by smaller and smaller positive values, and \(\frac{\pi}{x}\) increases without bound. When \(x\) is negative and approaching zero, \(\frac{\pi}{x}\) decreases without bound. In this sense, as we get close to \(x = 0\text{,}\) the inputs to the sine function are growing rapidly, and this leads to increasingly rapid oscillations in the graph of \(g\) between \(1\) and \(-1\text{.}\) If we plot the function \(g(x) = \sin\left(\frac{\pi}{x}\right)\) with a graphing utility and then zoom in on \(x = 0\text{,}\) we see that the function never settles down to a single value near the origin, which suggests that \(g\) does not have a limit at \(x = 0\text{.}\)

How do we reconcile the graph in Figure1.27 with Table1.26, which seems to suggest that the limit of \(g\) as \(x\) approaches \(0\) may in fact be \(0\text{?}\) The data misleads us because of the special nature of the sequence of input values \(\{0.1, 0.01, 0.001, \ldots\}\text{.}\) When we evaluate \(g(10^{-k})\text{,}\) we get \(g(10^{-k}) = \sin\left(\frac{\pi}{10^{-k}}\right) = \sin(10^k \pi) = 0\) for each positive integer value of \(k\text{.}\) But if we take a different sequence of values approaching zero, say \(\{0.3, 0.03, 0.003, \ldots\}\text{,}\) then we find that

That sequence of function values suggests that the value of the limit is \(\frac{\sqrt{3}}{2}\text{.}\) Clearly the function cannot have two different values for the limit, so \(g\) has no limit as \(x\) approaches \(0\text{.}\)

1. \((x^2 - 1)\) can be factored.
2. Expand the expression \((2+x)^3\text{,}\) and then combine like terms in the numerator.
3. Try multiplying the given function by this fancy form of 1: \(\frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}\text{.}\)

1. \(2\text{.}\)
2. \(12\text{.}\)
3. \(\frac{1}{2}\text{.}\)

Estimating the values of the limits with tables is straightforward and should suggest the exact values stated below.

1. \begin{align*} \lim_{x \to 1} \frac{x^2 - 1}{x-1} =\amp \lim_{x \to 1} \frac{(x+1)(x-1)}{x-1}\\ =\amp \lim_{x \to 1} (x+1) \\ =\amp 2\text{.} \end{align*}
2. \begin{align*} \lim_{x \to 0} \frac{(2+x)^3 - 8}{x} \amp = \lim_{x \to 0} \frac{8 + 12x + 6x^2 + x^3 - 8}{x}\\ \amp = \lim_{x \to 0} \frac{12x + 6x^2 + x^3}{x}\\ \amp = \lim_{x \to 0} (12 + 6x + x^2) \\ \amp = 12\text{.} \end{align*}
3. \begin{align*} \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \amp = \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}\\ \amp = \lim_{x \to 0} \frac{x+1-1}{x(\sqrt{x+1}+1)}\\ \amp = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1}\\ \amp = \frac{1}{2}\text{.} \end{align*}

1. Find the interval in which \(c\) lies and evaluate the function there.

2. Remember that for \(\lim_{x \to c^-} f(x)\text{,}\) we only consider values of \(x\) such that \(x \lt c\text{.}\) Find the appropriate formula to use in the piecewise definition for \(f\) to fit the values you are considering.

3. Use your work in (b) and compare left- and right-hand limits.

4. Use your work in (a) and (c).

5. Note that \(f\) is piecewise linear.

1. \(f(-2) = 1\text{;}\) \(f(-1)\) is not defined; \(f(0) = \frac{7}{3}\text{;}\) \(f(1) = 2\text{;}\) \(f(2) = 2\text{.}\)

2. \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3}\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3}\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2\text{.} \end{equation*}

3. \(\lim_{x \to -2} f(x)\) does not exist. The values of the limits as \(x \to c\) for \(c = -1, 0, 1, 2\) are \(\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}\)

4. \(c = -2\text{,}\) \(c = -1\text{,}\) and \(c = 1\text{.}\)

5. 

1. \(f(-2) = \frac{2}{3}(-2+2) + 1 = 1\text{;}\) \(f(-1)\) is not defined; \(f(0) = \frac{2}{3}(0+2)+1 = \frac{7}{3}\text{;}\) \(f(1) = 2\) (by the rule); \(f(2) = 4-2 = 2\text{.}\)

2. \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3}\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3}\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3\text{.} \end{equation*}
   \begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2\text{.} \end{equation*}

3. \(\lim_{x \to -2} f(x)\) does not exist because the left-hand limit is \(2\) while the right-hand limit is \(1\text{.}\) All of the other requested limits exist, as in each case the left- and right-hand limits exist and are equal. The respective values of the limits as \(x \to c\) for \(c = -1, 0, 1, 2\) are \(\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}\)

4. For \(c = -2\text{,}\) \(c = -1\text{,}\) and \(c = 1\text{,}\) \(\lim_{x \to c} f(x) \ne f(c)\text{.}\) At \(c = -2\text{,}\) the limit fails to exist, but \(f(-2) = 1\text{.}\) At \(c = -1\text{,}\) the limit is \(\frac{5}{3}\text{,}\) but \(f(-1)\) is not defined. At \(c = 1\text{,}\) the limit is 3, but \(f(1) = 2\text{.}\)

5. 

Consider evaluating limits on each side and comparing that value to the value of the function at the point.

1. \(f\) is continuous at \(x=2\text{.}\)
2. \(g\) is not continuous at \(x=2\text{.}\)
3. \(h\) is not continuous at \(x=2\text{.}\)

For each of these functions, we want to check that the limit exists at \(x=2\text{,}\) the function is defined at \(x=2\text{,}\) and these two values match.

1. We can examine the graph of \(y=f(x)\) at \(x=2\) or examine function values nearby \(x=2\) on the left and right to find that \(\lim_{x\to 2} f(x)=0\text{.}\) Evaluating \(f(2)=\ln(3-2)=\ln(1)=0\text{.}\) Thus, \(\lim_{x\to2}f(x)=f(2)\text{,}\) and \(f\) is continuous at \(x=2\text{.}\)
2. Notice that the graph of \(g\) has a vertical asymptote at \(x=2\text{,}\) so \(g(2)\) is undefined. Hence, \(g\) is not continuous at \(x=2\text{.}\)
3. For values of \(x\) near 2 (from the left and right), we have \(h(x)\) getting close to 5. Therefore, \(\lim_{x\to2}h(x)=5\text{.}\) However, \(h(2)=3\text{.}\) Since \(\lim_{x\to2}h(x)\neq h(2)\text{,}\) \(h\) is not continuous at \(x=2\text{.}\)

1. Consider the left- and right-hand limits at each value.

2. Carefully examine places on the graph where there's an open circle.

3. Are there locations on the graph where the function has a limit but there's a hole in the graph?

4. Remember that at least one of three conditions must fail: if the function lacks a limit, if the function is undefined, or if the limit exists but does not equal the function value, then \(f\) is not continuous at the point.

5. Note that the definition of being continuous requires the limit to exist.

1. \(c = -2\text{;}\) \(c = +2\text{.}\)

2. \(c = 3\text{.}\)

3. \(c = -1\text{;}\) \(c = 3\text{.}\)

4. \(c=-2\text{;}\) \(c = 2\text{;}\) \(c = 3\text{;}\) \(c = -1\text{.}\)

5. If \(f\) is continuous at \(x = c\text{,}\) then \(f\) has a limit at \(x = c\text{.}\)

1. \(\lim_{x \to c} f(x)\) does not exist at \(c = -2\) since

   \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ne -1 = \lim_{x \to -2^+}\text{,} \end{equation*}

   and \(\lim_{x \to c} f(x)\) does not exist at \(c = 2\) since \(\lim_{x \to 2^+} f(x)\) does not exist due to the infinitely oscillatory behavior of \(f\text{.}\)

2. The only point at which \(f\) is not defined is at \(c = 3\text{.}\)

3. At \(c = -1\text{,}\) note that \(\lim_{x \to -1} f(x)\) exists (and appears to have value approximately \(-3.25\)), but \(f(-1) = 1\) and thus \(\lim_{x \to -1} f(x) \ne f(-1)\text{.}\) At \(c = 3\text{,}\) we have \(\lim_{x \to 3} f(x) = -2.5\text{,}\) but \(f(3)\) is not defined so the limit exists but does not equal the function value.

4. Based on our work in (a), (b), and (c), \(f\) is not continuous at \(c=-2\) and \(c = 2\) because \(f\) does not have a limit at those points; \(f\) is not continuous at \(c = 3\) since \(f\) is not defined there; and \(f\) is not continuous at \(c = -1\) because at that point its limit does not equal its function value.

5. If \(f\) is continuous at \(x = c\text{,}\) then \(f\) has a limit at \(x = c\text{,}\) since one of the defining properties of being continuous at \(x = c\) is that the function has a limit at that input value. This shows that being continuous is a stronger condition than having a limit.