MA005: Calculus I

The precise ("formal") definition of limit carefully defines the ideas that we have already been using graphically and intuitively. The following side–by–side columns show some of the phrases we have been using to describe limits, and those phrases, particularly the last ones, provide the basis to building the definition of limit. 

Let's examine what the last phrase ("we can.."). means for the Particular Limit.

Example 1: We know $\lim\limits_{\mathrm{x} \rightarrow 3} 2 \mathrm{x}-1=5$. Show that we can guarantee that the values of $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-1$ are as close to 5 as we want by starting with values of $x$ sufficiently close to 3.

(a) What values of $x$ guarantee that $f(x)=2 x-1$ is within $\mathrm{1}$ unit of $\mathrm{5}$? (Fig. 1a)

Solution: "within $\mathrm{1}$ unit of $\mathrm{5}$" means between $5-1=4$ and $5+1=6$, so the question can be rephrased as "for what values of $x$ is $y=2 x-1$ between 4 and $6: 4 < 2 x-1 < 6$? We want to know which values of $x$ put the values of $y=2 x-1$ into the shaded band in Fig. 1a. The algebraic process is straightforward: solve $4 < 2 x-1 < 6$ for $x$ to get $5 < 2 x < 7$ and $2.5 < x < 3.5$. We can restate this result as follows: "If $\mathrm{x}$ is within $\mathbf{0. 5}$ units of 3, then $\mathrm{y}=2 \mathrm{x}-1$ is within $\mathrm{1}$ unit of $\mathrm{5}$". (Fig. 1b) 

Any smaller distance also satisfies the guarantee: e.g., "If $x$ is within $0.4$ units of 3, then $y=2 x-1$ is within 1 unit of 5". (Fig. 1c)

(b) What values of $x$ guarantee the $f(x)=2 x-1$ is within $0.2$ units of $5$? (Fig. 2a)

Solution: "within $0.2$ units of 5" means between $5-0.2=4.8$ and $5+0.2=5.2$, so the question can be rephrased as "for what values of $x$ is $y=2 x-1$ between $4.8$ and 5.2: $4.8 < 2 x-1 < 5.2$?" Solving for $x$, we get $5.8 < 2 x$ $< 6.2$ and $2.9 < x < 3.1$. "If $x$ is within $0.1$ units of 3, then $y=2 x-1$ is within $0.2$ units of $5.$" (Fig. 2b) Any smaller distance also satisfies the guarantee.

Rather than redoing these calculations for every possible distance from 5, we can do the work once, generally:

(c) What values of $x$ guarantee that $f(x)=2 x-1$ is within $E$ units of 5? (Fig. 3a)

Solution: "within $\mathrm{E}$ unit of 5" means between $5-\mathrm{E}$ and $5+\mathrm{E}$, so the question is "for what values of $x$ is $y=2 x-1$ between $5-E$ and $5+\varepsilon: 5-\mathrm{E} < 2 \mathrm{x}-1 < 5+\mathrm{E}$?" Solving $5-\mathrm{E} < 2 \mathrm{x}-1 < 5+\mathrm{E}$ for $\mathrm{x}, \mathrm{we}$ get $6-\mathrm{E} < 2 \mathrm{x} < 6+\mathrm{E}$ and $3-\mathrm{E} / 2 < \mathrm{x} < 3+\mathrm{E} / 2$. "If $\mathrm{x}$ is within $\mathrm{E} / 2$ units of 3, then $y=2 x-1$ is within $E$ units of 5". (Fig. 3b) Any smaller distance also satisfies the guarantee.

Part (c) of Example 1 illustrates a little of the power of general solutions in mathematics. Rather than doing a new set of similar calculations every time someone demands that $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-1$ be within some given distance of 5, we did the calculations once. And then we can respond for any given distance. For the question "What values of $x$ guarantee that $f(x)=2 x-1$ is within $0.4,0.1$ and $0.006$ units of 5?", we can answer "If $\mathrm{x}$ is within $0.2(=0.4 / 2), 0.05(=0.1 / 2)$ and $0.003(=0.006 / 2)$ units of 3".

Practice 1: $\quad \lim\limits_{\mathrm{x} \rightarrow 2} 4 \mathrm{x}-5=3$. What values of $\mathrm{x}$ guarantee that $\mathrm{f}(\mathrm{x})=4 \mathrm{x}-5$ is within

(a) 1 unit of 3?

(b) 0.08 units of 3?

(c) E units of 3? (Fig. 4)

The same ideas work even if the graphs of the functions are not straight lines, but the calculations are more complicated.

Example 2: $\quad \lim\limits_{\mathrm{x} \rightarrow 2} \mathrm{x}^{2}=4.$ (a) What values of $\mathrm{x}$ guarantee that $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ is within $\mathrm{1}$ unit of $\mathrm{4}$? (b) Within $0.2$ units of $4$?

(Fig. 5a) State each answer in the form "If $\mathrm{x}$ is within _____ units of $\mathrm{2}$, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{1}$ (or $0.2)$ unit of $\mathrm{4}$".

Solution; (a) If $x^{2}$ is within $\mathrm{1}$ unit of $\mathrm{4}$, then $3 < x^{2} < 5$ so $\sqrt{3} < x <$ $\sqrt{5}$

or $1.732 < \mathrm{x} < 2.236$. The interval containing these $\mathrm{x}$ values extends from $2-\sqrt{3} \approx 0.268$ units to the left of $\mathrm{2}$ to $\sqrt{5}-2 \approx 0.236$ units to the right of 2. Since we want to specify a single distance on each side of $\mathrm{2}$, we can pick the smaller of the two distances, $0.236$. (Fig. 5b)

"If $x$ is within $\mathrm{0.236}$ units of $\mathrm{2}$, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{1}$ unit of $\mathrm{4}$".

(b) Similarly, if $x^{2}$ is within $0.2$ units of 4, then $3.8 < x^{2} < 4.2$ so $\sqrt{3.8} < \mathrm{x} < \sqrt{4.2}$ or $1.949 < \mathrm{x} < 2.049$. The interval containing these $x$ values extends from $2-\sqrt{3.8} \approx 0.051$ units to the left of 2 to $\sqrt{4.2}-2 \approx 0.049$ units to the right of 2. Again picking the smaller of the two distances, "If $x$ is within $\mathrm{0.049}$ units of $\mathrm{2}$, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{1}$ unit of $\mathrm{4}$".

The situation in Example 2 of different distances on the left and right sides is very common, and we always pick our single distance to be the smaller of the distances to the left and right. By using the smaller distance, we can be certain that if $\mathrm{x}$ is within that smaller distance on either side, then the value of $\mathrm{f}(\mathrm{x})$ is within the specified distance of the value of the limit.

Practice 2: $\quad \lim\limits_{x \rightarrow 9} \sqrt{x}=3$. What values of $x$ guarantee that $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ is within 1 unit of 3? Within $0.2$ units of 3? (Fig. 6) State each answer in the form.

"If $x$ is ______ within units of 2, then $\mathrm{f}(\mathrm{x})$ is within 1 (or 0.2) unit of 4".

The same ideas can also be used when the function and the specified distance are given graphically, and in that case we can give the answer graphically.

Example 3: In Fig. 7, $\lim\limits_{\mathrm{x} \rightarrow 2} \mathrm{f}(\mathrm{x})=3$. What values of $\mathrm{x}$ guarantee that $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is within $\mathrm{E}$ units (given graphically) of 3? State your answer in the form "If $\mathrm{x}$ is within _____ (show a distance D graphically) of 2, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{E}$ units of 3".

Solution: $\quad$ The solution process requires several steps as illustrated in Fig. 8:

i. Use the given distance $\mathrm{E}$ to find the values $3-\mathrm{E}$ and $3+\mathrm{E}$ on the $\mathrm{y}$-axis.

ii. Sketch the horizontal band which has its lower edge at $\mathrm{y}=3-\mathrm{E}$ and its upper edge at $\mathrm{y}=3+\mathrm{E}$.

iii. Find the first locations to the right and left of $\mathrm{x}=2$ where the graph of $y=f(x)$ crosses the lines $y=3-E$ and $y=3+E$ and at these locations draw vertical lines to the $\mathrm{x}$-axis.

iv. On the $\mathrm{x}$-axis, graphically determine the distance from $\mathrm{2}$ to the vertical line on the left (labeled $\left.\mathrm{D}_{\mathrm{L}}\right)$ and from 2 to the vertical line on the right (labeled $\left.\mathrm{D}_{\mathrm{R}}\right)$.

v. Let the length $D$ be the smaller of the lengths $D_{L}$ and $D_{R}$.

Practice 3: In Fig. 9, $\lim\limits_{\mathrm{x} \rightarrow 3} \mathrm{f}(\mathrm{x})=1.8$. What of $x$ guarantee that $y=f(x)$ is within E units of $1.8$?