L'Hospital's rule question
Ronald Bruck
Sullivan) wrote:
:Hello
:
:I'm new to this group and would appreciate any hints anyone might give
:me for the following calculus problem:
:
:f(x)=1/2 (sin x - cos x sin x)
:g(x) = 1/2 (x - sin x)
:f(x)/g(x) is undefined, use f'(x)/g'(x) to find the limit of the ratio
:of areas as x approaches 0.
:According to approximations and my instructor, this should be 3.
:
:I calculated f'(x) = 1/2(cos x - cos^2 x + sin^2)
:and g'(x) = 1/2 (1 - cos x)
:
:I still get an undefined function as x > 0. My instructor hinted that
:I use the trig identities to make the limit of f'(x)/g'(x) as x > 0 =
:3. I've tried every manipulation I can come up with and no luck.
:
:Anyone wish to give me a hint as to which trig identity I should be
:using? Thanks in advance
What do others think of this kind of problem? Yes, we can make a simple
trig substitution (hint: it involves the Pythagorean theorem ;~), factor
out the 1 - cos x, and the answer drops out. A more clever student will
apply L'Hopital a SECOND time.
A REALLY clever student, or perhaps one who is just systematic, will use
the power-series representations for sin x and cos x -- order 3 is good
enough -- and write down the answer by inspection (esp. if he sees that
cos x sin x is 1/2 sin 2x). This technique has the advantage that it
always works for limits involving analytic functions. (I believe this is
the way Maple and Mathematica find such limits.)
Perhaps the MOST clever student will use the fact that cos x sin x = 1/2
sin 2x and divide numerator and denominator by x, then use (sin x)/x -->
1. What percentage do you think will do it that way? 1%? 5%?
I'm of two minds about this problem. It's a little better than routine,
and that's good. But it's milking a technique which I think is a dead
horse, namely L'Hopital's Rule.
--Ron Bruck
Now 100% ISDN from this address
Gunnar Johnsson
>Hello
>
>I'm new to this group and would appreciate any hints anyone might give
>me for the following calculus problem:
>
>f(x)=1/2 (sin x - cos x sin x)
>g(x) = 1/2 (x - sin x)
>f(x)/g(x) is undefined, use f'(x)/g'(x) to find the limit of the ratio
>of areas as x approaches 0.
>According to approximations and my instructor, this should be 3.
>
>I calculated f'(x) = 1/2(cos x - cos^2 x + sin^2)
>and g'(x) = 1/2 (1 - cos x)
>
>I still get an undefined function as x > 0. My instructor hinted that
>I use the trig identities to make the limit of f'(x)/g'(x) as x > 0 =
>3. I've tried every manipulation I can come up with and no luck.
>
>Anyone wish to give me a hint as to which trig identity I should be
>using? Thanks in advance
>
>tar...@ix.netcom.com
>
>Tara Sullivan
>tar...@ix.netcom.com
>or sull...@sonoma.edu
>
Suggestion: f'(x)=(using sin^2 x=1-cos^2 x) =1/2(cos x -2cos^2 x + 1) =
=(factoring -2t^2 + t + 1 = (1-t)(2t+1) )=1/2(1-cosx)(2cos...)
and you can divide away (1-cosx). gj.
Tara Sullivan
Robert Johnson
Tara Sullivan <tar...@ix.netcom.com> wrote:
>f(x)=1/2 (sin x - cos x sin x)
>g(x) = 1/2 (x - sin x)
>f(x)/g(x) is undefined, use f'(x)/g'(x) to find the limit of the ratio
>of areas as x approaches 0.
>According to approximations and my instructor, this should be 3.
>
>I calculated f'(x) = 1/2(cos x - cos^2 x + sin^2)
>and g'(x) = 1/2 (1 - cos x)
>
>I still get an undefined function as x > 0. My instructor hinted that
>I use the trig identities to make the limit of f'(x)/g'(x) as x > 0 =
>3. I've tried every manipulation I can come up with and no luck.
Attempt 1:
Drop the factors of 1/2 in f and g since they cancel. Use the Taylor
series for sine and cosine:
f(x) sin(x) (1-cos(x))
---- = -----------------
g(x) x - sin(x)
(x - x^3/6 + ...)(x^2/2 - x^4/24 + ...)
= ---------------------------------------
x^3/6 - x^5/120 + ...
(1 - x^2/6 + ...)(1/2 - x^2/24 + ...)
= -------------------------------------
1/6 - x^2/120 +...
-> (1/2)/(1/6)
= 3.
Thus, after cancelling 3 factors of x we get a rational expression
that clearly tends to 3.
Attempt 2:
Take up where you left off.
2
f'(x) cos(x) (1-cos(x)) + sin (x)
----- = ---------------------------
g'(x) 1 - cos(x)
cos(x) (1-cos(x)) + (1-cos(x)) (1+cos(x))
= -----------------------------------------
1 - cos(x)
= 1 + 2 cos(x)
-> 3
Rob Johnson
Apple Computer, Inc.
rjoh...@apple.com
Bill Dubuque
Date: 19 Mar 1996 06:31:56 GMT
In article <4il4do$d...@cloner3.netcom.com>, tar...@ix.netcom.com (Tara
Sullivan) wrote:
:
:I'm new to this group and would appreciate any hints anyone might give
:me for the following calculus problem:
:
:f(x)=1/2 (sin x - cos x sin x)
:g(x) = 1/2 (x - sin x)
:f(x)/g(x) is undefined, use f'(x)/g'(x) to find the limit of the ratio
:of areas as x approaches 0. ...
...A REALLY clever student, or perhaps one who is just systematic, will use
the power-series representations for sin x and cos x -- order 3 is good
enough -- and write down the answer by inspection (esp. if he sees that
cos x sin x is 1/2 sin 2x). This technique has the advantage that it
always works for limits involving analytic functions. (I believe this is
the way Maple and Mathematica find such limits.)
Indeed, power series are an effective algorithmic technique for
computing limits and are not limited to analytic functions. At
essential singularities one can resort to expansions in Hardy
fields of higher rank. This gives a decision procedure for
calculating limits in such fields, as I discovered circa 1980.
To the best of my knowledge this algorithm has not been fully
implemented in any computer algebra system; thus such systems
usually resort to inefficient heuristics such as L'Hospital's
rule in the worse case. For those interested in the details of
the algorithm I refer to work of John Shackell (j...@ukc.ac.uk)
who independently discovered a similar algorithm and (unlike I)
found the time to publish it--see his paper "Growth estimates
for Exp-Log functions", the Journal of Symbolic Computation, 10
(1990), 611-632. The pioneer of modern research in Hardy fields
is Max Rosenlicht, followed by M. Boshernitzan and J. Shackell.
There is also interesting related work by logicians studying
exponential rings model-theoretically (e.g. Lou van den Dries,
Wilkie, Dahn, Macintyre, Richardson, Khovanskii, etc).
Here's an example of my Macsyma implementation of my algorithm
computing a limit suggested by Schroeppel (who--not believing
in the existence of an algorithm for limits--designed this
monstrous expression in an attempt to thwart the algorithm)
x
e x
- e - e - x + a
e + x x
e e
e e
(d1) e - e
(c2) taylor(d1,x,inf,1); /* taylor series of d1 at x=inf, 1 term */
x
2 a 2 a - x x 2 a e
a e e e - e e - e
(d2)/T/ e + ((----- + ---------- + ...) e + ----- + ...) e + ...
2 2 2
a
thus lim d1 = e
x -> oo
Don't try L'Hospital's rule on this problem!
-Bill