L'Hospital's Rule

Note : The following are all indeterminate forms : , , , , , , .

Suppose f and g are functions such that f(a) = g(a) = 0 and that f' and g' are both continuous at x = a (recall: f' and g' continuous at a means that lim(f'(x)) = f'(a) and lim(g'(x)) = g'(a)). Consider the following:

Suppose f and g are continuous at a and that f(a) = g(a) = 0, so that = more work. Then,

= -- why?

= if

= (why is there no more " if ")

= by the limit laws

= f'(a)/g'(a) by definition of the derivative

= lim(f'(x))/lim(g'(x)) since f' and g' are continuous at a.

= a special case of a Big Rule:

L'Hopital's Rule : If applying the limit laws to produces the indeterminate form " " or " " then = . Also applies is a is replaced by or (i.e. rule applies also to "limits at infinity").

e.g. Evaluate the limits by two different methods (old & new):

1) 2)

> Limit((x^2-1)/(x-1),x=1)=limit((x^2-1)/(x-1),x=1);Limit((x^2-2*x+1)/(1-3*x+2*x^2),x=infinity)=limit((x^2-2*x+1)/(1-3*x+2*x^2),x=infinity);

e.g.'s

3) 4)

> Limit(sin(3*x)/sin(6*x),x=0)=limit(sin(3*x)/sin(6*x),x=0);

> Limit(sec(x)-tan(x),x=Pi/2,left)=limit(sec(x)-tan(x),x=Pi/2,left);

e.g.

Investigate the behaviour of f(x) = xln(x) around zero, determine the x-intercepts and any points where there is a horizontal tangent. Sketch.

> f:=x->x*ln(x);

Behaviour "close to" zero: (Note: Domain = (0, ) so there is no behaviour to the left of zero)

> Limit(f(x),x=0,right)=limit(f(x),x=0,right);

=> Since f -> 0 as x -> 0 there is no vertical asymptote at zero

x-intercept:

> solve(f(x)=0);

=> Crosses the x-axis at x = 1

Horizontal tangents:

> fprime:=diff(f(x),x);x_coord:=solve(fprime=0);

> f(x_coord);

=> There is one horizontal tangent asnd it occurs at (1/e,-1/e).

> plot(x*ln(x),x=-1..3,y=-1..4,colour=black);

Indeterminate Powers: " " , " " , " "

If produces one of the indeterminate powers above then let and take the logarithm of both sides. This technique depends on the unproven fact that "the log of the limit is the limit of the log."

e.g. Find y = .Note the limit has the indeterminate form . Once you take the log of both sides and use the property of logs that then you will obtain the indeterminate form " " which can be put into the form to which l'Hospital's Rule can be applied. Once you find the limit you must realize that you have the logarithm of the limit, not the limit itself . Find the domain of the above expression and evaluate the limit at 0. Determine whether there is a horizontal asymptote.

Domain: Recall that fractional powers are only defined for positive bases, so the domain consists of all x such that 1 - 2x > 0.

> solve(1-2*x>0);

> expr:=(1-2*x)^(1/x);

> Limit(expr,x=0,right)=limit(expr,x=0,right);

> Limit(expr,x=-infinity)=limit(expr,x=-infinity);

=> y = 1 is a horizontal asymptote .

> plot({expr,1},x=-4..1,y=0..1,colour=black);

e.g.

Find .

> expr:=x^(1/x);Limit(x^(1/x),x=infinity)=limit(x^(1/x),x=infinity);

> plot({expr,1},x=0..30,numpoints=2000,colour=black);

e.g. Continuously compounding interest.

Recall: if $P is invested at an annual interest rate r , expressed as a decimal, and compounded n times per year then then balance, B , payable in t years will be . If n is allowed to increase without bound then we have what is called continuously compounding interest.

1) Find a formula to calculate coninuously compounding interest and 2) find the balance due on $5,000.00 that is invested for 10 years at 8.5% per annum if interest is a) compounded quarterly b) compounded continuously.

Solution:

1) Use the formula for discreetly compounding interest and find the limit as n approaches infinity.

> B:=P*(1+r/n)^(n*t);

> Limit(B,n=infinity)=limit(B,n=infinity);

=> If $P is invested at a rate of r per annum and interest is compunded continuously then .

2a) $5000 invested at 8.5% (so r = 0.085) for 10 years (so t = 10) and compunded quarterly (so n = 4):

> subs({P=5000,r=0.085,n=4,t=10},B);

=> $11,594.52 in 10 years.

2b) Same as a) except compounded continuously (so ):

> 5000*exp(1)^(0.085*10);

> evalf(%);

=> $11,698.23 if compounding is continuously. Does it make sense that the second answer is larger?