Exercise 1
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Consider the problem of evaluating
.
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To work with this rational function we need to be able to access the numerator and denominator separately.
| > | top := x^3-9*x^2-45*x-91; |
| > | bot := x^2-5*x-104; |
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When we need the rational function, it can be formed as follows:
| > | F := top/bot; |
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The first attempt when evaluating this limit is to plug
into the rational expression
| > | eval( F, x=13 ); |
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As expected, this does not work. Note that both numerator and denominator are zero at
:
| > | plot( [top,bot], x=-10..15 ); |
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This suggests that (
) should be a factor of both numerator and denominator:
| > | top1 := factor( top ); |
| > | bot1 := factor( bot ); |
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Thus, for all values of
, the original rational function can be simplified to
| > | F1 := top1/bot1; |
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There are no problems evaluating this expression at
:
| > | eval( F1, x=13 ); |
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We have shown that
.
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Next, consider
. We could begin exactly as above, but suppose the expression is entered in rational form:
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| > | G := (sqrt(25+3*x) - sqrt(25-2*x))/(x^2+5*x); |
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Our first attempt is, as usual, to plug
into the expression:
| > | eval( G, x=0 ); |
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Not unexpectedly, this does not work. A plot of the expression near
suggests that the limit exists and that its value should be
.
| > | plot( G, x=-2..2 ); |
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Let's see if we can confirm this. To work with this expression we need to get our hands on the numerator and denominator. We could enter these manually or we could use Maple's numer and denom commands.
| > | top := numer( G ); |
| > | bot := denom( G ); |
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The trick in problems like these is to rationalize the difference of two square roots. This can be done by multiplying both numerator and denominator by (
).
| > | top1 := top * (sqrt(25+3*x)+sqrt(25-2*x)); |
| > | bot1 := bot * (sqrt(25+3*x)+sqrt(25-2*x)); |
| > | G1 := top1/bot1; |
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This is exactly where we started. The problem is that Maple saw the common factor and simplified. We have to tell Maple to expand the products before forming the quotient.
| > | top2 := expand(top1); |
| > | bot2 := expand(bot1); |
| > | G2 := top2/bot2; |
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Notice that the numerator is now much simpler but that the denominator is a real mess. This is not a concern because we see a common factor of x that can be factored from both. Hopefully this will be enough to allow us to evaluate the limit.
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| > | top3 := factor( top2 ); |
| > | bot3 := factor( bot2 ); |
| > | G3 := top3/bot3; |
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| > | eval( G3, x=0 ); |
| > | simplify( % ); |
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Great! The analytic and graphical limits agree.
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Another method for approximating limits is to prepare a table of values of the function at points close to the limit point. For this example we might do something like the following:
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| > | c := 0; |
| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]; |
| > | lim_table := [ x, `G(x)` ], [ `---------`, `---------` ]: |
| > | for X in x_vals do |
| > | lim_table := lim_table, [ X, eval( G, x=X ) ]; |
| > | end do: |
| > | convert( [lim_table], matrix ); |
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Evaluate each of the following limits, or explain why it does not exist. For each problem, report the limit and support your answer with a convincing table of values and a plot. Note: Each limit that exists should be expressed as an exact integer, rational, or irrational number, not as a floating-point number.
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a)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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b)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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c)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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d)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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e)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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f)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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g)
| > | F := (1+x)^(1/x); |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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h)
| > | F := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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i)
| > | F := %?;c := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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j)
| > | F := %?;c := %?; |
| > | c := %?; |
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| > | plot( F, x = %? .. %? ); |
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| > | x_vals := [ c-0.1, c-0.01, c-0.001, c-0.0001, c-0.00001, c+0.00001, c+0.0001, c+0.001, c+0.01, c+0.1 ]: |
| > | body := convert( [seq( [ X, eval( F, x=X ) ], X=x_vals )], matrix ): |
| > | header := matrix( 2, 2, [ [ `x`, `F(x)` ], [ `---------`, `---------` ] ] ): |
| > | stackmatrix( header, body ); |
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![x[1] = 2](images/lab223.gif){kind=small}
and
![x[2] = 4](images/lab224.gif){kind=small}
can by first finding the points on the graph of the function with these values of x:
) and ( 4,
) are on the graph of this function. The slope between these two points is
![x[1]](images/lab227.gif){kind=small}
,
![y[1]](images/lab228.gif){kind=small}
) with slope
is
![y = y[1]+m*(x-x[1])](images/lab230.gif){kind=small}
. Thus, the secant line we seek is
![x[2]](images/lab231.gif){kind=small}
approaches
![x[1]](images/lab232.gif){kind=small}
. The following loop creates a sequence of secant line plots and a table of numerical slopes for values of
![x[2]](images/lab233.gif){kind=small}
that approach
![x[1]](images/lab234.gif){kind=small}
.
![x[2]](images/lab235.gif){kind=small}
is the tangent line at
![x[1]](images/lab236.gif){kind=small}
. That is
.
![x[2]](images/lab238.gif){kind=small}
values (
x2_vals
) to obtain an estimate of the (right-hand) limit of the slope of the secant lines that is accurate to three digits to the right of the decimal point. In your report, include a table containing the
![x[2]](images/lab239.gif){kind=small}
values and the slope of the corresponding secant line.
![x[2]](images/lab240.gif){kind=small}
values that provides an estimate of the (left-hand) limit of the slope of the secant lines that is accurate to three digits to the right of the decimal point. This should be done in a table as in (a).
has a tangent line at
. If there is no tangent line at this point, explain why not. If there is a tangent line at this point, what is the slope of the tangent line (accuarate to three digits to the right of the decimal point) and the equation of the tangent line?
has a tangent line at
. If there is no tangent line at this point, explain why not. If there is a tangent line at this point, what is the slope of the tangent line (accuarate to three digits to the right of the decimal point) and the equation of the tangent line?
has a tangent line at
. If there is no tangent line at this point, explain why not. If there is a tangent line at this point, what is the slope of the tangent line (accuarate to three digits to the right of the decimal point) and the equation of the tangent line?
has a tangent line at
. If there is no tangent line at this point, explain why not. If there is a tangent line at this point, what is the slope of the tangent line (accuarate to three digits to the right of the decimal point) and the equation of the tangent line?
has a tangent line at
. If there is no tangent line at this point, explain why not. If there is a tangent line at this point, what is the slope of the tangent line (accuarate to three digits to the right of the decimal point) and the equation of the tangent line?