Calculus I

Lesson 21: L'Hospital's Rule

Example 1
We find each of the following limits, first without Maple and then with Maple.

Plot a graph to visualize your answer.

> restart; with(plots):

Warning, the name changecoords has been redefined

a)

We see:

= = .

> limit((x-2)/(x^2 - 4), x = 2);

> aa:= plot((x-2)/(x^2 - 4), x = 0..3, color = red):

> ba:= plot([x,0.25, x = 0..2], color = magenta):

> ca:= plot([2,x,x = 0..0.25], color = magenta):

> display(aa,ba,ca);

b)

We see:

= .

> limit( (x^6 - 1)/(x^4 - 1) , x = 1);

> ab:= plot((x^6 - 1)/(x^4 - 1), x = 0..2, color = red):

> bb:= plot([x,3/2,x = 0..1], color = magenta):

> cb:= plot([1,x,x = 0..3/2], color = magenta):

> display((ab,bb,cb));

c)

We see:

= = 1.

> limit( (exp(x) - 1)/(sin(x)), x = 0);

> ac:= plot((exp(x) - 1)/(sin(x)), x = -1..-0.001, color = red):

> bc:= plot((exp(x) - 1)/(sin(x)), x = 0.001..1, color = red):

> cc:= plot([x,1,x = -0.5..0.5], color = magenta):

> display((ac,bc,cc));

d)

We see:

= = 0.

> limit( (ln(x)) / x , x = infinity);

> plot((ln(x)) / x , x = 1..100);

e)

We see:

=

= = 1 * 1 = 1.

> limit( ((sin(x))^2) / ( tan(x^2)), x = 0);

> ae:= plot(((sin(x))^2) / ( tan(x^2)), x = -1..-0.01, color = red):

> be:= plot(((sin(x))^2) / ( tan(x^2)), x = 0.01..1, color = red):

> display((ae,be));

f)

We see:

= .

> limit( (arctan(2*x))/(3*x) , x = 0);

> af:= plot((arctan(2*x))/(3*x), x = -1..-0.01, color = red):

> bf:= plot((arctan(2*x))/(3*x), x = 0.01..1, color = red):

> cf:= plot([x,2/3,x = -0.5..0.5], color = magenta):

> display((af,bf,cf));

g)

We see:

=

= = 0.

> limit( sqrt(x)* ln(x) , x = 0, right);

> plot(sqrt(x)* ln(x), x = 0.001..5);

h)

We see:

=

=

= .

> limit( x * exp(1/x) - x, x = infinity);

> plot(x * exp(1/x) - x, x = 1..100, color = red);

i)

Let . Then, .

We see:

=

= = -1 * 0 = 0.

Hence,

= .

> limit( x ^(sin(x)), x = 0, right);

> plot(x ^(sin(x)), x = 0.001..2, color = red);

j)

Let . Then, .

We see:

=

= .

Hence,

= .

> limit( (exp(x) + x)^(1/x), x = infinity);

> plot((exp(x) + x)^(1/x), x = 2..50, color = red);

k)

Let . Then, .

We see:

=

= .

Hence,

= .

> limit( ( 1 + 1/x)^(x^2), x = infinity);

> plot(( 1 + 1/x)^(x^2), x = 2..10, color = red);

Example 2

The diagram shows a sector of a circle with central angle theta = .

Let A = A(theta) be the area of the region between the chord PR and the arc PR.

Let B = B(theta) be the area of the triangle PQR.

Find Limit( A(theta)/B(theta), = 0, right). Plot A(theta)/B(theta) to visualize your answer.

From your plot what happens to A(theta)/B(theta) as goes to /2?

Here is the code for the diagram.

> data:= ([0,0], [0.5, sqrt(3)/2]):

> data1:= ([0.5, sqrt(3)/2], [1,0]):

> data2:= ([0.5,sqrt(3)/2], [0.5,0]):

> a:= plot( sqrt(1 - x^2), x = 0..1, color = red):

> b:= plot([data], color = orange):

> c:= plot([data1], color = orange):

> d:= plot([data2], color = orange):

> e:= textplot([.07,.05,`O`],color = magenta):

> f:= textplot([.54,.05,`Q`], color = magenta):

> g:= textplot([.9,.05,`R`], color = magenta):

> h:= textplot([.4,.8,`P`], color = magenta):

> l:= textplot([.25,.15,`theta`], color = magenta):

> m:= textplot([.65,.35,`B`], color = red):

> n:= textplot([.8,.5,`A`], color = red):

> o:= plot(sqrt(.04 - x^2), x = 0.1..0.2, color = red):

> display(a,b,c,d,e,f,g,h,l,m,n,o, scaling=constrained);

Let the radius of the circle be r. We have:

= area of entire sector - area of triangle OPR =

and

= |QR| |PQ|/2 = .

Hence,

=

=

=

= .

> A:= x -> (0.5) * x^2 * ( x - sin(x));

> B:= x -> (0.5) * x * ( 1 - cos(x)) * x * sin(x);

> az:=plot(A(x)/B(x), x = 0.01..Pi/2, color = red):

> bz:= plot([x,1/3, x = 0.01..Pi/2], color = magenta):

> display((az,bz));

> evalf( A(Pi/2) / B(Pi/2) );

From the plot we see that as goes to /2, A( )/B( ) goes to approximately .