1.2 Numerical Limits

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We find limits using numerical information.

1 Limit notation

Limits are the backbone of calculus. A limit tells us the end of an infinite process. For example, consider the following infinite sequence of numbers: $0.9, 0.99, 0.999, 0.9999, 0.99999, 0.999999, ...$ This infinite sequence of numbers is becoming arbitrarily close to the number 1, so we say the limit of the sequence is 1. In calculus, we will be concerned with limits involving functions. The inputs of the function will undergo an infinite process which will then correspond to an infinite process for the outputs. Our goal will be to determine the limit of the outputs. Suppose that $f$ represents a function of the input variable $x$ . We denote by $x \to c$ an infinite process where the inputs are becoming arbitrarily close to the value $c$ without ever reaching it. Then, we denote by $\lim _{x\to c} f(x)$ the limit of the outputs of the function $f$ as $x$ approaches $c$ . The possibilities for the limit of the outputs of $f$ are: a numerical value, $\infty$ , $-\infty$ , or that the limit does not exist. If the numerical value of the limit is $L$ , then as the input variable $x$ approaches the value $c$ , the outputs, $f(x)$ approach the value $L$ , and we would write

$\lim _{x\to c} f(x) = L.$

If the value of the limit is $\infty$ , then as $x \to c$ , the outputs $f(x)$ are increasing without bound, and we write $\lim _{x \to c} f(x) = \infty .$

Similarly, if the limit is $-\infty$ , that means the outputs are decreasing without bound, and we write $\lim _{x \to c} f(x) = -\infty .$

If none of the previous conclusions apply, we say that the limit does not exist, and we write $\lim _{x \to c} f(x) \;\; \text {DNE}.$

In addition to writing $x \to c$ , we can also write the expression $x \to c^-$ to indicate that $x$ is approaching $c$ from the left hand side and $x \to c^+$ to indicate that $x$ is approaching $c$ from the right hand side.

The expression $x \to c^-$ can be understood visually using the following diagram:

Note that $x \to c^-$ implies $x < c$ . Similarly, the expression $x \to c^+$ can be understood visually using the following diagram:

Note that $x \to c^+$ implies $x > c$ . We refer to the limits

$\lim _{x \to c^-} f(x) \quad \text {and} \quad \lim _{x \to c^+} f(x)$

as one-sided limits, and we refer to

$\lim _{x \to c} f(x)$

as a two-sided limit.

We can also let the inputs, $x$ , either increase or decrease without bound, denoted by $x \to \infty$ and $x \to -\infty$ respectively. These can be represented visually by the following diagrams:

We refer to the limits $\lim _{x \to \infty } f(x) \quad \text {and} \quad \lim _{x \to -\infty } f(x)$ as limits at infinity.

2 Finding limits numerically

In what follows, functions will be presented using formulas. We will determine the limit of a function by making an appropriate table of values.

example 1 Determine the following one-sided limit: $\lim _{x \to 3^{+}} (x^2 + 2).$ First, $x \to 3^{+}$ means that $x$ is approaching the number $3$ from the right, so $x > 3$ . To represent this numerically, consider the following values for $x$ : $3.1, 3.01, 3.001, 3.0001$ . These numbers are each greater than 3 and they are getting closer to 3.

Now, to determine the limit numerically, we will plug these four values of $x$ into the function $x^2 + 2$ and look for a pattern in the outputs.

Here is the analysis:
If $x = 3.1$ then $x^2 + 2 = 11.61$ .
If $x = 3.01$ then $x^2 + 2 = 11.0601$ .
If $x = 3.001$ then $x^2 + 2 = 11.006001$ .
If $x = 3.0001$ then $x^2 + 2 = 11.00060001$ .

We can summarize this information in a table:

$\begin {array}{ c | c | c | c | c } x & 3.1 & 3.01 & 3.001 & 3.0001 \\ \hline x^2 + 2 & 11.61 & 11.0601 & 11.006001 & 11.00060001 \end {array}$

The outputs appear to be approaching the number 11, leading us to conclude that

$\lim _{x \to 3^+} (x^2 + 2) = 11$

In this particular example, we could have arrived at the answer by simply plugging the number $x = 3$ into the function $x^2 + 2$ since $3^2 + 2 = 11$ .

In general, when plugging in the value that $x$ is approaching yields a number, that number is usually the correct answer.

(problem 1) Compute the limit by plugging in the terminal value. $\lim _{x \to -4} (x^2 -3x -1)$

The limit is $\answer {27}$ .

We now turn our attention to examples where the shortcut of “plugging in” does not work.

example 2 Determine the following one-sided limit: $\lim _{x \to 1^{+}} \frac {x^3 - 1}{x^2 -1}.$ If we plug $x = 1$ into the function, we get the fraction $0/0$ , which is called an indeterminate form. This means that the answer can be anything and so a deeper analysis is required.

Since $x \to 1^{+}$ we know that $x$ is approaching the number $1$ from the right side, so $x > 1$ . To represent this numerically, consider the following values for $x$ : $1.1, 1.01, 1.001, 1.0001, 1.00001$ . These numbers are each greater than 1 and they are getting closer to 1.

To determine the limit numerically, we will plug these five values of $x$ into the function $\frac {x^3 - 1}{x^2-1}$ and look for a pattern in the outputs.

Here is the analysis:
If $x = 1.1$ then $\displaystyle {\frac {x^3 - 1}{x^2-1}= 1.57619}$ .
If $x = 1.01$ then $\displaystyle {\frac {x^3 - 1}{x^2-1} = 1.507512}$ .
If $x = 1.001$ then $\displaystyle {\frac {x^3 - 1}{x^2-1}= 1.500751}$ .
If $x = 1.0001$ then $\displaystyle {\frac {x^3 - 1}{x^2-1}= 1.5000750}$ .
If $x = 1.00001$ then $\displaystyle {\frac {x^3 - 1}{x^2-1}= 1.5000075}$ .

We can summarize this information in a table:

$\begin {array}{ c | c | c | c | c | c} x & 1.1 & 1.01 & 1.001 & 1.0001 & 1.00001\\ \hline \frac {x^3 - 1}{x^2-1} & 1.57619 & 1.507512 & 1.500751 & 1.500075 & 1.5000075 \end {array}$

The outputs appear to be approaching the number 1.5, leading us to conclude that

$\lim _{x \to 1^+} \frac {x^3 - 1}{x^2-1} = 1.5.$

(problem 2) Fill in the table below with 5 decimal places of accuracy and use it to find the limit: $\lim _{x \to 2^+} \frac {x^3 - 8}{x^2 - 4}.$

$\begin {array}{ c | c | c } x & 2.1 & 2.01 \\ \hline \frac {x^3 - 8}{x^2 - 4} & \answer {3.07561} & \answer {3.00751} \end {array}$ $\begin {array}{ c | c | c } x & 2.001 & 2.0001 \\ \hline \frac {x^3 - 8}{x^2 - 4} & \answer {3.00075} & \answer {3.00008} \end {array}$

Now determine the limit: $\lim _{x \to 2^+} \frac {x^3 - 8}{x^2 - 4} = \answer {3}$

The next example comes from solving a tangent line problem.

example 3 Compute the following limit: $\lim _{x \to 0^{+}} \frac {e^x-1}{x}.$ In this example, $x$ is near $0$ and $x>0$ . The values of $x$ we will use to generate a table of values are: $0.1, 0.01, 0.001$ and $0.0001.$

$\begin {array}{ c | c | c | c | c } x & 0.1 & 0.01 & 0.001 & 0.0001 \\ \hline \frac {e^x - 1}{x} & 1.052 & 1.005 & 1.0005 & 1.00005 \end {array}$

Based on the numerical information provided in the table above, we conclude that

$\lim _{x \to 0^+} \frac {e^x - 1}{x} = 1.$

Observe that plugging $x=0$ into the function $\displaystyle {f(x)= \frac {e^x -1}{x}}$ yields the indeterminate form $\frac {0}{0}$ , which is not the answer we sought.

(problem 3a) Fill in the table below with 5 decimal places of accuracy and use it to find the limit: $\lim _{x \to 0^-} \frac {e^x - 1}{x}.$

$\begin {array}{ c | c | c } x & -0.1 & -0.01 \\ \hline \frac {e^x - 1}{x} & \answer {0.95163} & \answer {0.99502} \end {array}$ $\begin {array}{ c | c | c } x & -0.001 & -0.0001 \\ \hline \frac {e^x - 1}{x} & \answer {0.99950} & \answer {0.99995} \end {array}$

Now determine the limit: $\lim _{x \to 0^-} \frac {e^x - 1}{x} = \answer {1}$

Based on the example above and the result of this problem, determine the two-sided limit: $\lim _{x \to 0} \frac {e^x - 1}{x}.$

The limit is (DNE is a possibility) $\answer {1}$

(problem 3b) Fill in the table below with 5 decimal places of accuracy and use it to find the limit: $\lim _{x \to 0^+} \frac {e^{2x} - 1}{x}.$

$\begin {array}{ c | c | c } x & 0.1 & 0.01 \\ \hline \frac {e^{2x} - 1}{x} & \answer {2.21403} & \answer {2.02013} \end {array}$ $\begin {array}{ c | c | c } x & 0.001 & 0.0001 \\ \hline \frac {e^{2x} - 1}{x} & \answer {2.00200} & \answer {2.00020} \end {array}$

Now determine the limit: $\lim _{x \to 0^+} \frac {e^{2x} - 1}{x} = \answer {2}$

example 4 Compute the following limit: $\lim _{x \to 0^+} \frac {\sin x}{x}.$

First, note that the shortcut of plugging $x=0$ into the function $\displaystyle {f(x)= \frac {\sin x}{x}}$ yields the indeterminate form $\frac {0}{0}$ . Hence, to compute this one-sided limit, we consider the following values for $x$ : $0.1,\ 0.01,\ 0.001$ and $\ 0.0001$ . Plugging these values into the function, we generate the following table of values:

$\begin {array}{ c | c | c | c | c } x & 0.1 & 0.01 & 0.001 & 0.0001 \\ \hline \frac {\sin x}{x} & 0.9983 & 0.99998 & 0.9999998 & 0.999999998 \end {array}$

Based on this numerical evidence, it would be reasonable to guess that $\lim _{x \to 0^{+}} \frac {\sin x}{x} = 1.$

(problem 4a) Fill in the table below with 9 decimal places of accuracy and use it to find the limit: $\lim _{x \to 0^-} \frac {\sin x}{x}.$

$\begin {array}{ c | c | c } x & -0.1 & -0.01 \\ \hline \frac {\sin x}{x} & \answer {0.998334166} & \answer {0.999983333} \end {array}$ $\begin {array}{ c | c | c } x & -0.001 & -0.0001 \\ \hline \frac {\sin x}{x} & \answer {0.999999833} & \answer {0.999999998} \end {array}$

Now determine the limit: $\lim _{x \to 0^-} \frac {\sin x}{x} = \answer {1}$

Based on the example above and the result of this problem, determine the two-sided limit: $\lim _{x \to 0} \frac {\sin x}{x}.$

The limit is (DNE is a possibility) $\answer {1}$

(problem 4b) Fill in the table below with 8 decimal places of accuracy and use it to find the limit: $\lim _{x \to 0^+} \frac {\sin 3x}{x}.$

$\begin {array}{ c | c | c } x & 0.1 & 0.01 \\ \hline \frac {\sin 3x}{x} & \answer {2.95520207} & \answer {2.99955002} \end {array}$ $\begin {array}{ c | c | c } x & 0.001 & 0.0001 \\ \hline \frac {\sin 3x}{x} & \answer {2.99999550} & \answer {2.99999996} \end {array}$

Now determine the limit: $\lim _{x \to 0^+} \frac {\sin 3x}{x} = \answer {3}$

example 5 Consider the limit: $\lim _{x \to -2^-} \frac {3}{x+2}.$ First we observe that plugging in the value $x=-2$ gives $\frac {3}{0}$ which is undefined. Thus we are required to make a deeper analysis to determine the limit. Since the values of $x$ are less than $-2$ , we consider values for $x$ such as $-2.1, -2.01, -2.001,$ and $-2.0001$ to make the following table of values:

$\begin {array}{ c | c | c | c | c| c| c} x & -2.1 & -2.01 & -2.001 & -2.0001 & -2.00001 & -2.000001\\ \hline \frac {3}{x+2} & -30 & -300 & -3,000 & -30,000 & -300,000 & -3,000,000 \end {array}$

The numerical evidence suggests that as $x$ approaches $-2$ from the left, the values of $f(x) = \frac {3}{x+2}$ are decreasing without bound. We conclude that

$\lim _{x \to -2^-} \frac {3}{x+2} = -\infty .$

This result has geometric significance. It means that the line $x=-2$ is a vertical asymptote for the graph of the function $f(x) = \frac {3}{x+2}.$

(problem 5) Fill in the table below and use it to find the limit: $\lim _{x \to 3^+} \frac {x}{x-3}.$

$\begin {array}{ c | c | c } x & 3.1 & 3.01 \\ \hline \frac {x}{x-3} & \answer {31} & \answer {301} \end {array}$ $\begin {array}{ c | c | c } x & 3.001 & 3.0001 \\ \hline \frac {x}{x-3} & \answer {3001} & \answer {30001} \end {array}$

Now determine the limit (type infinity for $\infty$ and -infinity for $-\infty$ ): $\lim _{x \to 3^+} \frac {x}{x-3} = \answer {\infty }$

In the following example, we discuss limits as the input $\infty$ . If $x \to \infty$ then $x$ is increasing without bound and we can use very large numbers for $x$ in our table.

example 6 Consider the limit: $\lim _{x \to \infty } \frac {3x -2}{x + 3}.$

Since $x \to \infty$ , we will use powers of ten to generate large values of $x$ when constructing our table:

$\begin {array}{ c | c | c | c | c|c} x & 100 & 1{,}000 & 10{,}000 & 100{,}000 & 1{,}000{,}000\\ \hline \frac {3x-2}{x+3} & 2.8932 & 2.9890 & 2.9989 & 2.99989 & 2.999989 \end {array}$

The numerical evidence suggests that as $x$ approaches $\infty$ , that is, as $x$ increases without bound, the values of $\frac {3x -2}{x +3}$ are approaching the number $3$ . Hence,

$\lim _{x \to \infty } \frac {3x -2}{x +3} = 3.$

This result has geometric significance. It means that the line $y = 3$ is a horizontal asymptote for the graph of the function $f(x) = \frac {3x -2}{x +3}.$

(problem 6) Fill in the table below using fractions and use it to find the limit: $\lim _{x \to \infty } \frac {3x-2}{5x+6}.$

$\begin {array}{ c | c | c } x & 10 & 100 \\ \hline \frac {3x-2}{5x+6} & \answer {28/56} & \answer {298/506} \end {array}$ $\begin {array}{ c | c | c } x & 1000 & 10000 \\ \hline \frac {3x-2}{5x+6} & \answer {2998/5006} & \answer {29998/50006} \end {array}$

Now determine the limit: $\lim _{x \to \infty } \frac {3x-2}{5x+6} = \answer {3/5}$

Limits and the number $\bf e$

We now consider a famous example involving compound interest and the number $e$ . The compound interest formula says $A = P\left (1+\frac {r}{n}\right )^{nt}$ where $A$ represents the amount of money resulting from investing a principle $P$ at an annual rate $r$ with the interest compounded $n$ times per year for $t$ years. A basic fact about compound interest is that the more frequently the interest is compounded, the faster the amount of money will grow. So, a natural question is “what happens as $n \to \infty ?$ ”

example 7 If we let $P=1, r= 1$ and $t=1$ in the compound interest formula above, we get: $A = \left (1 + \frac {1}{n}\right )^n.$ We will find the limit of $A$ as $n \to \infty$ , i.e.,

$\lim _{n \to \infty } \left (1+\frac {1}{n}\right )^n.$

Let’s construct a table of values with large values of $n$ and look for a pattern in the outputs: $\begin {array}{ c | c | c | c | c|c} n & 100 & 1,000 & 10,000 & 100,000 & 1,000,000\\ \hline A & 2.70481 & 2.71692 & 2.71815 & 2.71827 & 2.71828 \end {array}$

What we can see from this table is that even with the number of compounding periods being as large as 1,000,000 per year, the principle of $1 will not even grow to $3 by the end of the year. So there appears to be a limit to the effect of increasing the number of compounding periods on the amount of money generated by compound interest at a rate of 100%. This limit is the famous number $e$ that we see in the exponential function $e^x$ and as the base of the natural logarithm, $\ln x$ .

In conclusion, we have discovered the famous limit

$\lim _{n \to \infty } \left (1+\frac {1}{n}\right )^n = e,$

which can be used to define the number $e$ .

The renowned Swiss mathematician Leonhard Euler (1707-1783) was the first to refer to this number as $e$ , standing for “exponential”. $e$ is an irrational number, like $\pi$ and $\sqrt 2$ and to fifteen decimal places it is:

$e = 2.718281828459045...$

(problem 7) Fill in the table below with 5 decimal places of accuracy and use it to find the limit: $\lim _{x \to -\infty } \left (1+\frac {1}{x}\right )^x.$

$\begin {array}{ c | c | c } x & -100 & -1,000 \\ \hline \left (1+\frac {1}{x}\right )^x & \answer {2.73200} & \answer {2.71964} \end {array}$ $\begin {array}{ c | c | c } x & -10,000 & -100,000 \\ \hline (1+\frac {1}{x})^x & \answer {2.71842} & \answer {2.71830} \end {array}$

Now determine the limit (the answer is a well known number, denoted by a single letter): $\lim _{x \to -\infty } (1+\frac {1}{x})^x = \answer {e}$

4 Limits and instantaneous velocity

Rectilinear motion is motion along a straight line. We will consider an object to be in motion along a number line to keep track of its location. We let the function $s(t)$ denote the position of the object at time $t$ . Then the displacement of the object over a time interval, $[t_1, t_2]$ is given by $s(t_2) - s(t_1)$ . If we divide the displacement by the duration of the time interval, we get the average velocity of the object over that time interval: $\text {average velocity} = \frac {s(t_2) - s(t_1)}{t_2 -t_1}.$

Our goal is to determine the instantaneous velocity of the object at a given time. To do this, we will consider time intervals of shorter and shorter duration.

example 8 Suppose the position of a falling object is given by $s(t) = 100-16t^2$ where $t$ is measured in seconds and $s(t)$ is measured in feet. Find the instantaneous velocity of the object at time $t = 2$ seconds.
The average velocity of the object over the time interval $[t,2]$ is given by $\text {ave vel} = \frac {s(2)-s(t)}{2-t} = \frac {36-s(t)}{2-t},$ since $s(2) = 100 - 16(2^2) = 36$ feet. To obtain the instantaneous velocity, we will look for a pattern in the average velocity as the time interval $[t, 2]$ gets shorter and shorter: $\begin {array}{ c | c | c | c | c } t & 1.9 & 1.99 & 1.999 & 1.9999 \\ \hline \text {ave vel} & -62.4 & -63.84 & -63.984 & -63.9984 \end {array}$ It appears that as the average velocities are approaching the value $-64$ ft/sec, as the time intervals get shorter and shorter (negative velocity just means the object is falling). Hence we conclude that the instantaneous velocity at time $t = 2$ seconds is $-64$ ft/sec. Moreover, in terms of limits, the instantaneous velocity, $v(t)$ , is a limit of average velocities: $v(2) = \lim _{t\to 2^-} \frac {s(2) - s(t)}{2-t}.$ Technically, in this example we only considered the left hand limit, $t\to 2^-$ . In the next problem, we will verify that the right hand limit gives the same value.

(problem 8) Suppose the position of a falling object is given by $s(t) = 100-16t^2$ where $t$ is measured in seconds and $s(t)$ is measured in feet. Find the value of the limit $\lim _{t\to 2^+} \frac {s(t) - s(2)}{t-2},$ by filling in the table below.

$\begin {array}{ c | c | c } t & 2.1 & 2.01 \\ \hline \text {ave vel} & \answer {-65.6} & \answer {-64.16} \end {array}$ $\begin {array}{ c | c | c } t & 2.001 & 2.0001 \\ \hline \text {ave vel} & \answer {-64.016} & \answer {-64.0016} \end {array}$

Now determine the instantaneous velocity as a limit of the average velocities: $v(2) = \lim _{t\to 2^+} \frac {s(t) - s(2)}{t-2} = \answer {-64} \text {ft/sec}.$

Here is a detailed, lecture style video on numerical limits:

2024-09-27 13:54:56

1 Limit notation

2 Finding limits numerically

3 Limits and the number \bf e

4 Limits and instantaneous velocity

Limits and the number $\bf e$