Exercise 1.1.5. PROTEUS EXERCISE.
How does the process of investigating the behavior of \(f(x)=\sin(x)/x\) near \(x=0\) differ from doing the same near \(x=1\text{?}\)
Section 1.1 An Introduction To Limits
We begin our study of limits by considering examples that demonstrate key concepts that will be explained as we progress.
Consider the function \(y = \frac{\sin(x) }{x}\text{.}\) When \(x\) is near the value 1, what value (if any) is \(y\) near?
While our question is not precisely formed (what constitutes “near the value 1”?), the answer does not seem difficult to find. One might think first to look at a graph of this function to approximate the appropriate \(y\) values. Consider Figure 1.1.3, where \(y = \frac{\sin(x) }{x}\) is graphed. For values of \(x\) near 1, it seems that \(y\) takes on values near \(0.85\text{.}\) In fact, when \(x=1\text{,}\) then \(y=\frac{\sin(1) }{1} \approx 0.84\text{,}\) so it makes sense that when \(x\) is “near” 1, \(y\) will be “near” \(0.84\text{.}\)
Graph of \(\sin(x)/x\text{,}\) shown for \(x\) between \(-7\) and \(7\text{,}\) and \(y\) between \(0\) and \(1\text{.}\) The \(x\) intercepts are at \(x=-2\pi, -\pi, \pi\text{,}\) and \(2\pi\text{,}\) and a \(y\) intercept is at \(y = 1\text{.}\) The graph has a downward curve for \(-\pi \lt x \lt \pi\) and an upward curve for \(-2\pi \lt x \lt -\pi\text{,}\) and \(\pi \lt x \lt 2\pi\text{.}\) The graph is undefined for \(x = 0\text{.}\)
Graph of \(\sin(x)/x\) zoomed in on values where \(x\) is near \(1\text{.}\) This view of the graph shows \(x\) from \(0.5\) to \(1.5\text{.}\) The graph has only a slight downward curve. It shows that for \(x = 1\text{,}\) \(\sin(x)/x\) is approximately \(0.84\)
Consider this same function again at a different value for \(x\text{.}\) When \(x\) is near \(0\text{,}\) what value (if any) is \(y\) near? By considering Figure 1.1.4, one can see that it seems that \(y\) takes on values near \(1\text{.}\) But what happens when \(x=0\text{?}\) We have
\begin{equation*}
y \rightarrow \frac{\sin(0) }{0} \rightarrow {\genfrac{}{}{0pt}{0}{\text{“}}{}}\frac{0}{0}{\genfrac{}{}{0pt}{0}{\text{”}}{}}\text{.}
\end{equation*}
The expression \(0/0\) has no value; it is indeterminate. Such an expression gives no information about what is going on with the function nearby. We cannot find out how \(y\) behaves near \(x=0\) for this function simply by letting \(x=0\text{.}\)
Graph of \(\sin(x)/x\) zoomed in on values where \(x\) is near \(0\text{.}\) The image shows the portion of the graph where \(x\) is from \(-1\) to \(1\text{.}\) The graph has a downward curve and is symmetric about \(x=0\text{.}\) The height of the graph approaches \(y = 1\) when \(x\) is near \(0\text{.}\) A hollow dot at the point \((0,1)\) shows that the function is undefined when \(x = 0\text{;}\) that is, \(f(0) =\) undefined.
Finding a limit entails understanding how a function behaves near a particular value of \(x\text{.}\) Before continuing, it will be useful to establish some notation. Let \(y=f(x)\text{;}\) that is, let \(y\) be a function of \(x\) for some function \(f\text{.}\) The expression “the limit of \(y\) as \(x\) approaches \(1\)” describes a number, often referred to as \(L\text{,}\) that \(y\) nears as \(x\) nears \(1\text{.}\) We write all this as
\begin{equation*}
\lim_{x\to 1} y = \lim_{x\to 1} f(x) = L\text{.}
\end{equation*}
This is not a complete definition (that will come in the next section); this is a pseudo-definition that will allow us to explore the idea of a limit.
Above, where \(f(x) = \sin(x)/x\text{,}\) we approximated
\begin{equation*}
\lim_{x\to 1} \frac{\sin(x) }{x} \approx 0.84 \quad \text{ and } \quad \lim_{x\to 0}\frac{\sin(x) }{x} \approx 1\text{.}
\end{equation*}
(We approximated these limits, hence used the “\(\approx\)” symbol, since we are working with the pseudo-definition of a limit, not the actual definition.)
Once we have the true definition of a limit, we will find limits analytically; that is, determining exact values using a variety of mathematical tools. For now, we will approximate limits both graphically and numerically. Graphing a function can provide a good approximation, though often not very precise. Numerical methods can provide a more accurate approximation. We have already approximated limits graphically, so we now turn our attention to numerical approximations.
Consider again \(\lim_{x\to 1}\frac{\sin(x)}{x}\text{.}\) To approximate this limit numerically, we can create a table of \(x\) and \(f(x)\) values where \(x\) is “near” \(1\text{.}\) This is done in Figure 1.1.7.
Notice that for values of \(x\) near \(1\text{,}\) we have \(\sin(x)/x\) near \(0.841\text{.}\) The \(x=1\) row is included, but we stress the fact that when considering limits, we are not concerned with the value of the function at that particular \(x\) value; we are only concerned with the values of the function when \(x\) is near 1.
| \(x\) | \(\sin(x)/x\) |
| 0.9 | 0.870363 |
| 0.99 | 0.844471 |
| 0.999 | 0.841772 |
| 1 | 0.841471 |
| 1.001 | 0.841170 |
| 1.01 | 0.838447 |
| 1.1 | 0.810189 |
Now approximate \(\lim_{x\to 0} \frac{\sin(x)}{x}\) numerically. We already approximated the value of this limit as \(1\) graphically in Figure 1.1.4. Figure 1.1.8 shows the value of \(\sin(x)/x\) for values of \(x\) near \(0\text{.}\) Ten places after the decimal point are shown to highlight how close to \(1\) the value of \(\sin(x)/x\) gets as \(x\) takes on values very near \(0\text{.}\) We include the \(x=0\) row but again stress that we are not concerned with the value of our function at \(x=0\text{,}\) only on the behavior of the function near \(0\text{.}\)
| \(x\) | \(\sin(x)/x\) |
| -0.1 | 0.9983341665 |
| -0.01 | 0.9999833334 |
| -0.001 | 0.9999998333 |
| 0 | not defined |
| 0.001 | 0.9999998333 |
| 0.01 | 0.9999833334 |
| 0.1 | 0.9983341665 |
This numerical method gives confidence to say that \(1\) is a good approximation of \(\lim_{x\to 0} \frac{\sin(x)}{x}\text{;}\) that is,
\begin{equation*}
\lim_{x\to 0} \frac{\sin(x)}{x} \approx 1\text{.}
\end{equation*}
Later we will be able to prove that the limit is exactly \(1\text{.}\)
Exercise 1.1.9. PROTEUS EXERCISE.
Why does it make sense to talk about the limit \(\lim_{x\to 0}\dfrac{\sin(x)}{x}\) if \(\dfrac{\sin(0)}{0}\) is not defined?
We now consider several examples that allow us explore different aspects of the limit concept.
Example 1.1.10. Approximating the value of a limit.
Use graphical and numerical methods to approximate
\begin{equation*}
\lim_{x\to 3} \frac{x^2-x-6}{6x^2-19x+3}\text{.}
\end{equation*}
Solution 1.
To graphically approximate the limit, graph
\begin{equation*}
y = \frac{x^2-x-6}{6x^2-19x+3}
\end{equation*}
on a small interval that contains \(3\text{.}\) To numerically approximate the limit, create a table of values where the \(x\) values are near \(3\text{.}\) This is done in Figure 1.1.11 and Figure 1.1.12, respectively.
Graph of \(f(x)=\frac{x^2 - x - 6}{6x^2 - 19x + 3}\text{,}\) zoomed on values near \(x = 3\text{,}\) and showing the portion of the graph for \(x\) from \(2.5\) to \(3.5\text{.}\)
There is a slight upward curve to the graph. The graph suggests that the limit of the function as \(x\) approaches \(3\) is \(0.294\text{.}\) The graph also shows that the function is undefined for \(x = 3\text{.}\)
| \(x\) | \(\frac{x^2-x-6}{6x^2-19x+3}\) |
| \(2.9\) | \(0.29878\) |
| \(2.99\) | \(0.294569\) |
| \(2.999\) | \(0.294163\) |
| 3 | not defined |
| \(3.001\) | \(0.294073\) |
| \(3.01\) | \(0.293669\) |
| \(3.1\) | \(0.289773\) |
The graph shows that when \(x\) is near \(3\text{,}\) the value of \(y\) is very near \(0.3\text{.}\) By considering values of \(x\) near \(3\text{,}\) we see that \(y=0.294\) is a better approximation. The graph and the table imply that
\begin{equation*}
\lim_{x\to 3} \frac{x^2-x-6}{6x^2-19x+3} \approx 0.294\text{.}
\end{equation*}
Solution 2. Video solution
This example may bring up a few questions about approximating limits (and the nature of limits themselves).
-
If a graph does not produce as good an approximation as a table, why bother with it?
-
How many values of \(x\) in a table are “enough?” In the previous example, could we have just used \(x=3.001\) and found a fine approximation?
Graphs are useful since they give a visual understanding concerning the behavior of a function. Sometimes a function may act “erratically” near certain \(x\) values which is hard to discern numerically but very plain graphically (see Example 1.1.24). Since graphing utilities are very accessible, it makes sense to make proper use of them.
Since tables and graphs are used only to approximate the value of a limit, there is not a firm answer to how many data points are “enough.” Include enough so that a trend is clear, and use values (when possible) both less than and greater than the value in question. In Example 1.1.10, we used both values less than and greater than \(3\text{.}\) Had we used just \(x=3.001\text{,}\) we might have been tempted to conclude that the limit had a value of \(0.3\text{.}\) While this is not far off, we could do better. Using values “on both sides of 3” helps us identify trends.
Example 1.1.13. Approximating the value of a limit.
Graphically and numerically approximate the limit of \(f(x)\) as \(x\) approaches \(0\text{,}\) where
\begin{equation*}
f(x) = \begin{cases}x+1 \amp x\lt 0 \\ -x^2+1 \amp x \gt 0\end{cases}\text{.}
\end{equation*}
Solution 1.
Again we graph \(f(x)\) and create a table of its values near \(x=0\) to approximate the limit. Note that this is a piecewise defined function, so it behaves differently on either side of \(0\text{.}\) Figure 1.1.14 shows a graph of \(f(x)\text{,}\) and on either side of \(0\) it seems the \(y\) values approach \(1\text{.}\) Note that \(f(0)\) is not actually defined, as indicated in the graph with the open circle.
Graph of the piecewise-defined function in Example 1.1.13. For values of \(x \lt 0\) the graph is straight with a slope of \(1\) and for values of \(x \gt 0\) the graph curves downward. A hollow dot at the point \((0,1)\) shows that at \(x = 0\text{,}\) \(f(x)\) is undefined. However, both parts of the graph, for \(x\lt 0\) and for \(x\gt 0\text{,}\) get close to the point \((0,1)\) as \(x\) gets close to \(0\text{.}\)
| \(x\) | \(f(x)\) |
| \(-0.1\) | \(0.9\) |
| \(-0.01\) | \(0.99\) |
| \(-0.001\) | \(0.999\) |
| \(0.001\) | \(0.999999\) |
| \(0.01\) | \(0.9999\) |
| \(0.1\) | \(0.99\) |
Figure 1.1.15 shows values of \(f(x)\) for values of \(x\) near \(0\text{.}\) It is clear that as \(x\) takes on values very near \(0\text{,}\) \(f(x)\) takes on values very near \(1\text{.}\) It turns out that if we let \(x=0\) for either “piece” of \(f(x)\text{,}\) \(1\) is returned; this is significant and we’ll return to this idea later.
The graph and table allow us to say that \(\lim_{x\to 0}f(x) \approx 1\text{;}\) in fact, we are probably very sure it equals 1.
Solution 2. Video solution
Identifying When Limits Do Not Exist.
A function may not have a limit for all values of \(x\text{.}\) That is, we cannot write that \(\lim_{x\to c}f(x)=L\) (where \(L\) is some real number) for all values of \(c\text{,}\) for there may not be a number that \(f(x)\) is approaching. There are three common ways in which a limit may fail to exist.
We’ll explore each of these in turn.
Example 1.1.17. Different Values Approached From Left and Right.
Explore why \(\lim_{x\to 1} f(x)\) does not exist, where
\begin{equation*}
f(x) = \begin{cases}x^2-2x+3 \amp x\leq 1 \\ x \amp x \gt 1\end{cases}\text{.}
\end{equation*}
Solution.
A graph of \(f(x)\) around \(x=1\) and a table are given in Figures Figure 1.1.18 and Figure 1.1.19, respectively. It is clear that as \(x\) approaches \(1\text{,}\) \(f(x)\) does not seem to approach a single number. Instead, it seems as though \(f(x)\) approaches two different numbers. When considering values of \(x\) less than \(1\) (approaching \(1\) from the left), it seems that \(f(x)\) is approaching \(2\text{;}\) when considering values of \(x\) greater than \(1\) (approaching \(1\) from the right), it seems that \(f(x)\) is approaching \(1\text{.}\) Recognizing this behavior is important; we’ll study this in greater depth later. Right now, it suffices to say that the limit does not exist since \(f(x)\) is approaching two different values as \(x\) approaches \(1\text{.}\)
Graph of piecewise function in Example 1.1.17. For values of \(x \leq 1\) the graph has a upward curve, and the graph ends at the point \((1,2)\text{,}\) illustrating the fact that \(f(1)=2\text{.}\)
For values of \(x \gt 1\) the graph is a straight line with a positive slope. Moving left to right, the line begins at the point \((1,1)\text{,}\) at which there is a hollow dot, indicating that to the right of \(x=1\text{,}\) the value of \(f(x)\) approaches 1.
The most important feature of the graph is that it shows how \(f(x)\) approaches two different values as \(x\) approaches \(1\text{,}\) depending on whether \(x\lt 1\) or \(x\gt 1\text{.}\)
| \(x\) | \(f(x)\) |
| \(0.9\) | \(2.01\) |
| \(0.99\) | \(2.0001\) |
| \(0.999\) | \(2.000001\) |
| \(1.001\) | \(1.001\) |
| \(1.01\) | \(1.01\) |
| \(1.1\) | \(1.1\) |
Example 1.1.20. The Function Grows Without Bound.
Explore why \(\lim_{x\to 1} \frac{1}{(x-1)^2}\) does not exist.
Solution.
A graph and table of \(f(x) = \frac{1}{(x-1)^2}\) are given in Figure 1.1.21 and Figure 1.1.22, respectively. Both show that as \(x\) approaches \(1\text{,}\) \(f(x)\) grows larger and larger.
Graph of the function for Example 1.1.20. The graph hows a horizontal asymptote at \(y = 0\) and a vertical asymptote at \(x = 1\text{.}\) Because of the vertical asymptote at \(x = 1\) the function has no limit as \(x\) approaches \(1\text{.}\)
| \(x\) | \(f(x)\) |
| \(0.9\) | \(100\text{.}\) |
| \(0.99\) | \(10000\text{.}\) |
| \(0.999\) | \(1.\times 10^6\) |
| \(1.001\) | \(1.\times 10^6\) |
| \(1.01\) | \(10000\text{.}\) |
| \(1.1\) | \(100\text{.}\) |
We can deduce this on our own, without the aid of the graph and table. If \(x\) is near 1, then \((x-1)^2\) is very small, and:
\begin{equation*}
\frac{1}{\text{ very small number } } = \text{ very large number }\text{.}
\end{equation*}
Since \(f(x)\) is not approaching a single number, we conclude that
\begin{equation*}
\lim_{x\to 1}\frac{1}{(x-1)^2}
\end{equation*}
does not exist.
Example 1.1.24. The Function Oscillates.
Explore why \(\lim_{x\to 0}\sin(1/x)\) does not exist.
Solution 1.
Two graphs of \(f(x) = \sin(1/x)\) are given in Figure 1.1.25. Figure 1.1.25.(a) shows \(f(x)\) on the interval \([-1,1]\text{;}\) notice how \(f(x)\) seems to oscillate near \(x=0\text{.}\) One might think that despite the oscillation, as \(x\) approaches \(0\text{,}\) \(f(x)\) approaches \(0\text{.}\) However, Figure 1.1.25.(b) zooms in on \(\sin(1/x)\text{,}\) on the interval \([-0.1,0.1]\text{.}\) Here the oscillation is even more pronounced. Finally, in Figure 1.1.26, we see \(\sin(1/x)\) evaluated for values of \(x\) near \(0\text{.}\) As \(x\) approaches \(0\text{,}\) \(f(x)\) does not appear to approach any value.
The graph of \(f(x)=sin(1/x)\) is shown, for \(x\) values between \(-1\) and \(1\text{.}\) Like any sinusoidal graph, the curve oscillates back and forth between \(y=1\) and \(y=-1\text{.}\) However, as \(x\) gets close to \(0\text{,}\) the argument of the sine function increases rapidly, causing the distance between successive peaks to get smaller and smaller as the graph nears the \(y\) axis. As \(x\) gets close to zero, the oscillations get so close together that it is no longer possible to distinguish them, and the curve appears to become a solid, vertical strip.
Another graph of \(f(x)=\sin(1/x)\) is shown, this time zoomed in to show only the \(x\) interval from \(-0.1\) to \(0.1\text{.}\) The features of the graph are the similar to what is visible over the larger interval: further from the origin, we see the graph oscillating (rapidly) between \(y=1\) and \(y=-1\text{.}\) Near the orgin, the oscillations become so rapid that we can no longer tell them apart. What we conclude from the graph is that on any interval containing \(x=0\text{,}\) \(f(x)=\sin(1/x)\) takes on every \(y\) value between \(-1\) and \(1\text{.}\) (In fact, \(f(x)\) attains every value infinitely many times!)
| \(x\) | \(\sin(1/x)\) |
| \(0.1\) | \(-0.544021\) |
| \(0.01\) | \(-0.506366\) |
| \(0.001\) | \(0.82688\) |
| \(0.0001\) | \(-0.305614\) |
| \(1.\times 10^{-5}\) | \(0.0357488\) |
| \(1.\times 10^{-6}\) | \(-0.349994\) |
| \(1.\times 10^{-7}\) | \(0.420548\) |
It can be shown that in reality, as \(x\) approaches 0, \(\sin(1/x)\) takes on all values between \(-1\) and \(1\) infinitely many times! Because of this oscillation, \(\lim_{x\to 0}\sin(1/x)\) does not exist.
Solution 2. Video solution
Exercise 1.1.27. PROTEUS EXERCISE.
Limits of Difference Quotients.
We have approximated limits of functions as \(x\) approached a particular number. We will consider another important kind of limit after explaining a few key ideas.
Let \(f(x)\) represent the position function, in feet, of some particle that is moving in a straight line, where \(x\) is measured in seconds. Let’s say that when \(x=1\text{,}\) the particle is at position \(10\) ft., and when \(x=5\text{,}\) the particle is at \(20\) ft. Another way of expressing this is to say
Since the particle traveled \(10\) feet in \(4\) seconds, we can say the particle’s average velocity was \(2.5\) ft/s. We write this calculation using a “quotient of differences,” or, a difference quotient:
\begin{equation*}
\frac{f(5) - f(1)}{5-1} \frac{\text{ ft}}{\text{s}} = \frac{10 \text{ ft}}{4\text{ s}} = 2.5 \text{ ft/s }\text{.}
\end{equation*}
This difference quotient can be thought of as the familiar “rise over run” used to compute the slopes of lines. In fact, that is essentially what we are doing: given two points on the graph of \(f\text{,}\) we are finding the slope of the secant line through those two points. See Figure 1.1.29.
The image shows the graph of a function, along with a line that intersects the graph at two points. The graph has the shape of a parabola that opens downward, and is displayed over the region \(0\leq x\leq 6\text{,}\) with a \(y\) range from 0 to 25. There are two points plotted on the graph at coordinates \((1, 10)\) and \((5, 20)\text{,}\) and the line through these points is an example of a secant line.
Now consider finding the average speed on another time interval. We again start at \(x=1\text{,}\) but consider the position of the particle \(h\) seconds later. That is, consider the positions of the particle when \(x=1\) and when \(x=1+h\text{.}\) The difference quotient (excluding units) is now
\begin{equation*}
\frac{f(1+h)-f(1)}{(1+h)-1} = \frac{f(1+h)-f(1)}{h}\text{.}
\end{equation*}
Let \(f(x) = -1.5x^2+11.5x\text{;}\) note that \(f(1)=10\) and \(f(5) = 20\text{,}\) as in our discussion. We can compute this difference quotient for all values of \(h\) (even negative values!) except \(h=0\text{,}\) for then we get “0/0,” the indeterminate form introduced earlier. For all values \(h\neq 0\text{,}\) the difference quotient computes the average velocity of the particle over an interval of time of length \(h\) starting at \(x=1\text{.}\)
For small values of \(h\text{,}\) i.e., values of \(h\) close to \(0\text{,}\) we get average velocities over very short time periods and compute secant lines over small intervals. See Figure 1.1.30. This leads us to wonder what the limit of the difference quotient is as \(h\) approaches \(0\text{.}\) That is,
\begin{equation*}
\lim_{h\to 0} \frac{f(1+h)-f(1)}{h} = \text{ ? }
\end{equation*}
Graph of the function from Figure 1.1.29, with the points on the graph \((1,10)\) and \((3,21)\) marked. A secant line is drawn through these points; it has a steeper slope than in Figure 1.1.29. Here the value of \(h\) is \(2\text{.}\)
Graph of the function from Figure 1.1.30.(a), but with the points \((1,10)\) and \((2,17)\) on the graph marked. These points correspond to a value of \(h=1\text{,}\) and the secant line through these points has a steeper slope than in Figure 1.1.30.(a).
Graph of the function from Figure 1.1.30.(b), but with the points \((1,10)\) and \((1.5,13.875)\) on the graph marked, corresponding to the value \(h=0.5\text{.}\) The secant line through these points again has a steeper slope than in the previous figures.
As we do not yet have a true definition of a limit nor an exact method for computing it, we settle for approximating the value. While we could graph the difference quotient (where the \(x\)-axis would represent \(h\) values and the \(y\)-axis would represent values of the difference quotient) we settle for making a table. See Figure 1.1.31. The table gives us reason to assume the value of the limit is about \(8.5\text{.}\)
| \(h\) | \(\frac{f(1+h)-f(1)}{h}\) |
| \(-0.5\) | \(9.25\) |
| \(-0.1\) | \(8.65\) |
| \(-0.01\) | \(8.515\) |
| \(0.01\) | \(8.485\) |
| \(0.1\) | \(8.35\) |
| \(0.5\) | \(7.75\) |
Proper understanding of limits is key to understanding calculus. With limits, we can accomplish seemingly impossible mathematical things, like adding up an infinite number of numbers (and not get infinity) and finding the slope of a line between two points, where the “two points” are actually the same point. These are not just mathematical curiosities; they allow us to link position, velocity and acceleration together, connect cross-sectional areas to volume, find the work done by a variable force, and much more.
Exercise 1.1.33. PROTEUS EXERCISE.
In some calculus textbooks, a “definition” of the limit is given as follows:
Let \(f(x)\) be a function defined on an open interval containing \(c\text{,}\) but not possibly at \(c\text{.}\) We say that the limit of \(f(x)\) as \(x\) approaches \(c\) is \(L\text{,}\) and write\begin{equation*} \lim_{x\to c}f(x)=L\text{,} \end{equation*}if \(f(x)\) can be made arbitrarily close to \(L\) by choosing \(x\) sufficiently close, but not equal, to \(c\text{.}\)
Can you think of any shortcomings with this definition? Why wouldn’t mathematicians consider this definition to be sufficient? (In particular, why are we about to consider a more “precise” definition of the limit in the next section?)
In the next section we give the formal definition of the limit and begin our study of finding limits analytically. In the following exercises, we continue our introduction and approximate the value of limits.
Exercises Exercises
Terms and Concepts
1.
2.
3.
True.
-
The limit considers values of \(x\) near to \(5\text{,}\) but not equal to \(5\text{.}\)The case where the limit is equal to \(f(5)\) has a special name; we will consider this in detail in Section 1.5.
False.
-
The limit considers values of \(x\) near to \(5\text{,}\) but not equal to \(5\text{.}\)The case where the limit is equal to \(f(5)\) has a special name; we will consider this in detail in Section 1.5.
4.
Describe three situations where \(\lim\limits_{x\to c}f(x)\) does not exist.
5.
In your own words, what is a difference quotient?
6.
Problems
Exercise Group.
Approximate the limit numerically and graphically.
7.
\(\lim\limits_{x\to 1}\left({x^{2}+3x-5}\right)\)
8.
\(\lim\limits_{x\to -1}\left({x^{3}-x^{2}+x+1}\right)\)
9.
\(\lim\limits_{x\to 0}\left({\frac{x+2}{x^{2}+5x}}\right)\)
10.
\(\lim\limits_{x\to -1}\left({\frac{x^{2}-2x-3}{x^{2}+3x+2}}\right)\)
11.
\(\lim\limits_{x\to 1}\left({\frac{x^{2}-9x+8}{x^{2}+x-2}}\right)\)
12.
\(\lim\limits_{x\to 3}\left({\frac{x^{2}+13x-27}{x^{2}-6x+9}}\right)\)
13.
\(\lim\limits_{x\to 3}f(x)\text{,}\) where \(f(x)={\begin{cases}\displaystyle{x+2}\amp \text{if}\ x \le 3\cr
\displaystyle{2x+1}\amp \text{if}\ x > 3\end{cases}}\)
14.
\(\lim\limits_{x\to 3}f(x)\text{,}\) where \(f(x)={\begin{cases}\displaystyle{x^{2}-3x-4}\amp \text{if}\ x \le 3\cr
\displaystyle{3x-13}\amp \text{if}\ x > 3\end{cases}}\)
15.
\(\lim\limits_{x\to 0}f(x)\text{,}\) where \(f(x)={\begin{cases}\displaystyle{\cos\mathopen{}\left(x\right)}\amp \text{if}\ x \le 0\cr
\displaystyle{x^{2}+5x+1}\amp \text{if}\ x > 0\end{cases}}\)
16.
\(\lim\limits_{x\to {\frac{\pi }{2}}}f(x)\text{,}\) where \(f(x)=\begin{cases}\sin(x)\amp x\leq {\frac{\pi }{2}}\\\cos(x)\amp x\gt {\frac{\pi }{2}}\end{cases}\)
17.
\(\lim\limits_{x\to 0}\lvert x\rvert^x\)
18.
\(\lim\limits_{x\to0}e^{-e^{1/x}}\)
19.
\(\lim\limits_{x\to-2}\big\lfloor\lvert x\rvert\big\rfloor !\text{,}\) where \(\lvert x\rvert\) is the absolute value of \(x\text{,}\) \(\lfloor x\rfloor\) is the floor of \(x\) (the greatest integer less than or equal to \(x\)), and \(x!\) is \(x\) factorial.
20.
\(\lim\limits_{x\to0}\big\lfloor\lvert x\rvert\big\rfloor !\text{,}\) where \(\lvert x\rvert\) is the absolute value of \(x\text{,}\) \(\lfloor x\rfloor\) is the floor of \(x\) (the greatest integer less than or equal to \(x\)), and \(x!\) is \(x\) factorial.
Exercise Group.
Approximate the limit of the difference quotient, \(\lim\limits_{h\to 0}\frac{f(a+h)-f(a)}{h}\text{,}\) using \(h = \pm 0.1, \pm 0.01\text{.}\)
21.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
22.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
23.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
24.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
25.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
26.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
27.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
28.
| \(h\) | \(\frac{f(a+h)-f(a)}{h}\) |
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