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

\(x\) |
\(f(x)\) |

\(-0.9\) |
\(2.9\) |

\(-0.99\) |
\(2.99\) |

\(-0.999\) |
\(2.999\) |

\(-0.9999\) |
\(2.9999\) |

\(-1.1\) |
\(3.1\) |

\(-1.01\) |
\(3.01\) |

\(-1.001\) |
\(3.001\) |

\(-1.0001\) |
\(3.0001\) |

\(x\) |
\(f(x)\) |

\(-1.9\) |
\(3.9\) |

\(-1.99\) |
\(3.99\) |

\(-1.999\) |
\(3.999\) |

\(-1.9999\) |
\(3.9999\) |

\(-2.1\) |
\(4.1\) |

\(-2.01\) |
\(4.01\) |

\(-2.001\) |
\(4.001\) |

\(-2.0001\) |
\(4.0001\) |

Table1.21Table of \(f\) values near \(x=-1\text{.}\)Table1.22Table of \(f\) values near \(x=-2\text{.}\)Figure1.23Plot of \(y=f(x)\) on \([-4,2]\text{.}\)

From Table1.21, 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 Figure1.23. 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 Table1.22 and Figure1.23, it appears that \(f\) should have a limit of \(4\) at \(x = -2\text{.}\)

To see algebraically why this is the case, observe that

\begin{align*}
\lim_{x \to -2} f(x) = \amp \lim_{x \to -2} \frac{4-x^2}{x+2}\\
= \amp \lim_{x \to -2} \frac{(2-x)(2+x)}{x+2}\text{.}
\end{align*}

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

\begin{equation*}
\lim_{x \to -2} f(x) = \lim_{x \to -2} (2-x)\text{.}
\end{equation*}

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.

\(x\) |
\(g(x)\) |

\(2.9\) |
\(0.84864\) |

\(2.99\) |
\(0.86428\) |

\(2.999\) |
\(0.86585\) |

\(2.9999\) |
\(0.86601\) |

\(3.1\) |
\(0.88351\) |

\(3.01\) |
\(0.86777\) |

\(3.001\) |
\(0.86620\) |

\(3.0001\) |
\(0.86604\) |

\(x\) |
\(g(x)\) |

\(-0.1\) |
\(0\) |

\(-0.01\) |
\(0\) |

\(-0.001\) |
\(0\) |

\(-0.0001\) |
\(0\) |

\(0.1\) |
\(0\) |

\(0.01\) |
\(0\) |

\(0.001\) |
\(0\) |

\(0.0001\) |
\(0\) |

Table1.24Table of \(g\) values near \(x=3\text{.}\)Table1.25Table of \(g\) values near \(x=0\text{.}\)Figure1.26Plot of \(y=g(x)\) on \([-4,4]\text{.}\)

First, as \(x\) approaches \(3\text{,}\) it appears from the values in Table1.24 that the function is approaching a number between \(0.86601\) and \(0.86604\text{.}\) From the graph in Figure1.26 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

\begin{equation*}
\lim_{x \to 3} g(x) = \frac{\sqrt{3}}{2}\text{.}
\end{equation*}

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 Figure1.26 with Table1.25, 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

\begin{equation*}
g(3 \cdot 10^{-k}) = \sin\left(\frac{\pi}{3 \cdot 10^{-k}}\right) = \sin\left(\frac{10^k \pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866025\text{.}
\end{equation*}

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 \(0\text{.}\)