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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Table 1.2.4 Table of \(f\) values near \(x=-1\text{.}\)Table 1.2.5 Table of \(f\) values near \(x=-2\text{.}\)Figure 1.2.6 Plot of \(f(x)\) on \([-4,2]\text{.}\)

From Table 1.2.4, 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 \(f\text{.}\) To see this a bit more rigorously and from an algebraic point of view, consider the formula for \(f\text{:}\) \(f(x) = \frac{4-x^2}{x+2}\text{.}\) The numerator and denominator are each polynomial functions, which are among the most well-behaved functions that exist. Formally, such functions are *continuous*^{ 2 }, which means that the limit of the function at any point is equal to its function value. Here, it follows that as \(x \to -1\text{,}\) \((4-x^2) \to (4 - (-1)^2) = 3\text{,}\) and \((x+2) \to (-1 + 2) = 1\text{,}\) so as \(x \to -1\text{,}\) the numerator of \(f\) tends to 3 and the denominator tends to 1, hence \(\lim_{x \to -1} f(x) = \frac{3}{1} = 3\text{.}\)

The situation is more complicated when \(x \to -2\text{,}\) due in part to the fact that \(f(-2)\) is not defined. If we attempt to use a similar algebraic argument regarding the numerator and denominator, we observe that as \(x \to -2\text{,}\) \((4-x^2) \to (4 - (-2)^2) = 0\text{,}\) and \((x+2) \to (-2 + 2) = 0\text{,}\) so as \(x \to -2\text{,}\) the numerator of \(f\) tends to 0 and the denominator tends to 0. We call \(0/0\) an *indeterminate form* and will revisit several important issues surrounding such quantities later in the course. For now, we simply observe that this tells us there is somehow more work to do. From Table 1.2.5 and Figure 1.2.6, it appears that \(f\) should have a limit of \(4\) at \(x = -2\text{.}\) To see algebraically why this is the case, let's work directly with the form of \(f(x)\text{.}\) 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*}

At this point, it is important to observe that since we are taking the limit as \(x \to -2\text{,}\) 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 now find that

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Table 1.2.7 Table of \(g\) values near \(x=3\text{.}\)Table 1.2.8 Table of \(g\) values near \(x=0\text{.}\)Figure 1.2.9 Plot of \(g(x)\) on \([-4,4]\text{.}\)

First, as \(x \to 3\text{,}\) it appears from the data (and the graph) that the function is approaching approximately \(0.866025\text{.}\) To be precise, we have to use the fact that \(\frac{\pi}{x} \to \frac{\pi}{3}\text{,}\) and thus we find that \(g(x) = \sin(\frac{\pi}{x}) \to \sin(\frac{\pi}{3})\) as \(x \to 3\text{.}\) The exact value of \(\sin(\frac{\pi}{3})\) is \(\frac{\sqrt{3}}{2}\text{,}\) which is approximately 0.8660254038. Thus, we see that

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

As \(x \to 0\text{,}\) 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 wild oscillations in the graph of \(g\text{.}\) It is an instructive exercise to plot the function \(g(x) = \sin\left(\frac{\pi}{x}\right)\) with a graphing utility and then zoom in on \(x = 0\text{.}\) Doing so shows that the function never settles down to a single value near the origin and suggests that \(g\) does not have a limit at \(x = 0\text{.}\)

How do we reconcile this with the righthand table above, which seems to suggest that the limit of \(g\) as \(x\) approaches \(0\) may in fact be \(0\text{?}\) Here we need to recognize that the data misleads us because of the special nature of the sequence \(\{0.1, 0.01, 0.001, \ldots\}\text{:}\) 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 data would suggest that the value of the limit is \(\frac{\sqrt{3}}{2}\text{.}\) Clearly the function cannot have two different values for the limit, and this shows that \(g\) has no limit as \(x \to 0\text{.}\)