1.2 Numerical Limits

$\newenvironment {prompt}{}{} \newcommand {\bart }[0]{BART} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We find limits using numerical information.

Numerical Limits

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

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 $x = 3.001$ 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.

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.

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 $x$ : $1.1, 1.01, 1.001, 1.0001$ . These numbers are each greater than 1 and they are getting closer to 1.

To determine the limit numerically, we will plug these four 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 $\frac {x^3 - 1}{x^2-1}= 1.57619$ .
If $x = 1.01$ then $\frac {x^3 - 1}{x^2-1} = 1.507512$ .
If $x = 1.001$ then $\frac {x^3 - 1}{x^2-1}= 1.500751$ .
If $x = 1.0001$ then $\frac {x^3 - 1}{x^2-1}= 1.5000750$ .
If $x = 1.00001$ then $\frac {x^3 - 1}{x^2-1}= 1.5000075$ .

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

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

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: $\lim _{x \to 2^+} \frac {x^3 - 8}{x^2 - 4} = \answer [given]{3}$

Compute the following one-sided limit: $\lim _{x \to 3^{-}} \frac {|x-3|}{x-3}.$ In this limit, $x$ is approaching $3$ from the left, so $x<3$ . We therefore consider the following numerical values for $x$ : $2.9, 2.99, 2.999, 2.9999$ . These values are all less than three and they are getting closer to $3$ . Plugging these values into the function $|x-3|/(x-3)$ yields the following table of values:

$\begin {array}{ c | c | c | c | c } x & 2.9 & 2.99 & 2.999 & 2.9999 \\ \hline \frac {|x-3|}{x-3} & -1 & -1 & -1 & -1 \end {array}$

The values of the function are are all equal to $-1$ , and so we can conclude that $\lim _{x \to 3^{-}} \frac {|x-3|}{x-3} = -1.$ It should be noticed that the shortcut of plugging in $x=3$ yields the meaningless indeterminate form $\frac 00$ .

Fill in the table below with 5 decimal places of accuracy and use it to find the limit: $\lim _{x \to -3^-} \frac {2x+6}{|x+3|}.$

$\begin {array}{ c | c | c } x & -3.1 & -3.01 \\ \hline \frac {2x+6}{|x+3|} & \answer {-2} & \answer {-2} \end {array}$ $\begin {array}{ c | c | c } x & -3.001 & -3.0001 \\ \hline \frac {2x+6}{|x+3|} & \answer {-2} & \answer {-2} \end {array}$

Now determine the limit: $\lim _{x \to -3^-} \frac {2x+6}{|x+3|} = \answer [given]{-2}$

The next example comes from solving a tangent line problem.

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.

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

$\begin {array}{ c | c | c } x & -0.1 & -0.01 \\ \hline \frac {e^x - 1}{x} & \answer {0.95163} & \answer {0.99502} \end {array}$ $\begin {array}{ c | c | c } x & -0.001 & -0.0001 \\ \hline \frac {e^x - 1}{x} & \answer {0.9995} & \answer {0.99995} \end {array}$

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

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) $\answer {1}$

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

$\begin {array}{ c | c | c } x & 0.1 & 0.01 \\ \hline \frac {e^{2x} - 1}{x} & \answer {2.21403} & \answer {2.02013} \end {array}$ $\begin {array}{ c | c | c } x & 0.001 & 0.0001 \\ \hline \frac {e^{2x} - 1}{x} & \answer {2.002} & \answer {2.0002} \end {array}$

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

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 $\lim _{x \to 0^{+}} \frac {\sin (x)}{x} = 1.$

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

$\begin {array}{ c | c | c } x & -0.1 & -0.01 \\ \hline \frac {\sin (x)}{x} & \answer {0.9983} & \answer {0.99998} \end {array}$ $\begin {array}{ c | c | c } x & -0.001 & -0.0001 \\ \hline \frac {\sin (x)}{x} & \answer {0.999998} & \answer {0.9999998} \end {array}$

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

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) $\answer {1}$

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

$\begin {array}{ c | c | c } x & 0.1 & 0.01 \\ \hline \frac {\sin (3x)}{x} & \answer {2.9552} & \answer {2.99955} \end {array}$ $\begin {array}{ c | c | c } x & 0.001 & 0.0001 \\ \hline \frac {\sin (3x)}{x} & \answer {3} & \answer {3} \end {array}$

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

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 $f(x) = \frac {2}{x-3}$ 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 $f(x) = \frac {3}{x+2}.$

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 $-\infty$ ): $\lim _{x \to 3^+} \frac {x}{x-3} = \answer [given]{\infty }$

Consider the limit: $\lim _{x \to \frac {\pi }{2}^-} \tan (x).$ Since $\tan (x) = \frac {\sin (x)}{\cos (x)},$ $\sin \big (\frac {\pi }{2}\big ) = 1 \ \text {and} \ \cos \big (\frac {\pi }{2}\big ) = 0$ we have $\tan \big (\frac {\pi }{2}\big ) = \frac {1}{0}$ which is undefined, so a deeper analysis is required to determine the limit. To understand the behavior of the tangent function for values of $x$ near $\frac {\pi }{2}$ , use a calculator to make a table of values noting that $\frac {\pi }{2} \approx 1.570796327$ .

$\begin {array}{ c | c | c | c | c | c} x & 1.5 & 1.57 & 1.5707 & 1.57079 & 1.570796 \\ \hline \tan (x) & 14.101 & 1,255.77 & 10,381.33 & 158,057.91 & 3,060,023.31 \end {array}$

The numerical evidence suggests that as $x$ approaches $\frac {\pi }{2}$ from the left, the values of $f(x) = \tan (x)$ are increasing without bound. Therefore, we are led to conclude that

$\lim _{x \to \frac {\pi }{2}^-} \tan (x) = \infty .$

This result has geometric significance. It means that the graph of the function $f(x) = \tan (x)$ has a vertical asymptote at $x=\frac {\pi }{2}.$

Fill in the table below with two decimal places of accuracy and use it to find the limit: $\lim _{x \to \frac {\pi }{2}^+} \tan (x).$

$\begin {array}{ c | c | c } x & 1.6 & 1.58 \\ \hline \tan (x) & \answer {-34.23} & \answer {-108.65} \end {array}$ $\begin {array}{ c | c | c } x & 1.571 & 1.5708 \\ \hline \tan (x) & \answer {-4909.83} & \answer {-272241.81} \end {array}$

Now determine the limit (type infinity for $\infty$ and -infinity for $-\infty$ ): $\lim _{x \to \frac {\pi }{2}^+} \tan (x) = \answer [given]{-\infty }$

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

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 $f(x) = \frac {3x -2}{x +3}.$

Consider the limit: $\lim _{x \to -\infty } \frac {2x + 1}{3x -4}.$

Since $x \to -\infty$ , we will use negative powers of ten when constructing our table:

$\begin {array}{ c | c | c | c | c |c} x & -100 & -1{,}000 & -10{,}000 & -100{,}000 & -1{,}000{,}000\\ \hline \frac {2x + 1}{3x -4} & .6546 & .6654 & .6665 & .66665 & .666665 \end {array}$

The numerical evidence suggests that as $x$ approaches $-\infty$ , that is, as $x$ decreases without bound, the values of $\frac {2x + 1}{3x -4}$ are approaching the decimal $0.6666...$ which we recognize as the fraction $2/3$ . Hence,

$\lim _{x \to -\infty } \frac {2x + 1}{3x -4} = \frac {2}{3}.$

This result has geometric significance. It means that the line $y = 2/3$ is a horizontal asymptote for the graph of the function $f(x) = \frac {2x + 1}{3x -4}.$

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: $\lim _{x \to \infty } \frac {3x-2}{5x+6} = \answer [given]{3/5}$

Compound Interest 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 $n \to \infty ?$ ” To answer this question, we will simplify matters by letting the principle, $P$ be $1$ , the interest rate be $100\$ (so that $r = 1$ ) and $t = 1$ year. In this case, $A = \left (1+\frac {1}{n}\right )^n,$

and we will try to find $\lim _{n \to \infty } \left (1+\frac {1}{n}\right )^n.$

To do this, let’s construct a table of values with large numbers for $n$ :

$\begin {array}{ c | c | c | c | c|c} n & 100 & 1,000 & 10,000 & 100,000 & 1,000,000\\ \hline A & 2.70481 & 2.71692 & 2.71815 & 2.71827 & 2.71828 \end {array}$

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

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

This limit is one way to define the number $e$ .

This number was coined by the renowned Swiss mathematician Leonhard Euler in the eighteenth century. He used the symbol $e$ to stand for ”exponential” and to fifteen decimal places, the (irrational) number $e$ is:

$e = 2.718281828459045...$

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): $\lim _{x \to -\infty } (1+\frac {1}{x})^x = \answer [given]{e}$

Rectilinear Motion 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: $\text {average velocity} = \frac {s(t_2) - s(t_1)}{t_2 -t_1}.$

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 average velocity of the object over the time interval $[1.5, 2]$ .

A negative velocity indicates that the object is falling.

The average velocity is $\answer {-56}$ ft/sec.

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.

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 is a limit of average velocities: $\text {inst vel} = \lim _{t\to 2} \frac {s(2) - s(t)}{2-t}.$ 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.

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: $\lim _{t\to 2^+} \frac {s(t) - s(2)}{t-2} = \answer [given]{-64} \text {ft/sec}.$

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