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@settitle The limit procedure

@deffn {library procedure} limit proc x1 x2 k
@deffnx {library procedure} limit proc x1 x2

@var{Proc} must be a procedure taking a single inexact real argument.
@var{K} is the number of points on which @var{proc} will be called; it
defaults to 8.

If @var{x1} is finite, then @var{Proc} must be continuous on the
half-open interval:

 ( @var{x1} .. @var{x1}+@var{x2} ]

And @var{x2} should be chosen small enough so that @var{proc} is
expected to be monotonic or constant on arguments between @var{x1} and
@var{x1} + @var{x2}.

@code{Limit} computes the limit of @var{proc} as its argument
approaches @var{x1} from @var{x1} + @var{x2}.
@code{Limit} returns a real number or real infinity or @samp{#f}.

If @var{x1} is not finite, then @var{x2} must be a finite nonzero real
with the same sign as @var{x1}; in which case @code{limit} returns:

@code{(limit (lambda (x) (proc (/ x))) 0.0 (/ @var{x2}) @var{k})}

@code{Limit} examines the magnitudes of the differences between
successive values returned by @var{proc} called with a succession of
numbers from @var{x1}+@var{x2}/@var{k} to @var{x1}.

If the magnitudes of differences are monotonically decreasing, then
then the limit is extrapolated from the degree n polynomial passing
through the samples returned by @var{proc}.

If the magnitudes of differences are increasing as fast or faster than
a hyperbola matching at @var{x1}+@var{x2}, then a real infinity with
sign the same as the differences is returned.

If the magnitudes of differences are increasing more slowly than the
hyperbola matching at @var{x1}+@var{x2}, then the limit is
extrapolated from the quadratic passing through the three samples
closest to @var{x1}.

If the magnitudes of differences are not monotonic or are not
completely within one of the above categories, then #f is returned.
@end deffn

@example
;; constant
(limit (lambda (x) (/ x x)) 0 1.0e-9)           ==> 1.0
(limit (lambda (x) (expt 0 x)) 0 1.0e-9)        ==> 0.0
(limit (lambda (x) (expt 0 x)) 0 -1.0e-9)       ==> +inf.0
;; linear
(limit + 0 976.5625e-6)                         ==> 0.0
(limit - 0 976.5625e-6)                         ==> 0.0
;; vertical point of inflection
(limit sqrt 0 1.0e-18)                          ==> 0.0
(limit (lambda (x) (* x (log x))) 0 1.0e-9)     ==> -102.70578127633066e-12
(limit (lambda (x) (/ x (log x))) 0 1.0e-9)     ==> 96.12123142321669e-15
;; limits tending to infinity
(limit + +inf.0 1.0e9)                          ==> +inf.0
(limit + -inf.0 -1.0e9)                         ==> -inf.0
(limit / 0 1.0e-9)                              ==> +inf.0
(limit / 0 -1.0e-9)                             ==> -inf.0
(limit (lambda (x) (/ (log x) x)) 0 1.0e-9)     ==> -inf.0
(limit (lambda (x) (/ (magnitude (log x)) x)) 0 -1.0e-9)
                                                ==> -inf.0
;; limit doesn't exist
(limit sin +inf.0 1.0e9)                        ==> #f
(limit (lambda (x) (sin (/ x))) 0 1.0e-9)       ==> #f
(limit (lambda (x) (sin (/ x))) 0 -1.0e-9)      ==> #f
(limit (lambda (x) (/ (log x) x)) 0 -1.0e-9)    ==> #f
;; conditionally convergent - return #f
(limit (lambda (x) (/ (sin x) x)) +inf.0 1.0e222)
                                                ==> #f
;; asymptotes
(limit / -inf.0 -1.0e222)                       ==> 0.0
(limit / +inf.0 1.0e222)                        ==> 0.0
(limit (lambda (x) (expt x x)) 0 1.0e-18)       ==> 1.0
(limit (lambda (x) (sin (/ x))) +inf.0 1.0e222) ==> 0.0
(limit (lambda (x) (/ (+ (exp (/ x)) 1))) 0 1.0e-9)
                                                ==> 0.0
(limit (lambda (x) (/ (+ (exp (/ x)) 1))) 0 -1.0e-9)
                                                ==> 1.0
(limit (lambda (x) (real-part (expt (tan x) (cos x)))) (/ pi 2) 1.0e-9)
                                                ==> 1.0
;; This example from the 1979 Macsyma manual grows so rapidly
;;  that x2 must be less than 41.  It correctly returns e^2.
(limit (lambda (x) (expt (+ x (exp x) (exp (* 2 x))) (/ x))) +inf.0 40)
                                                ==> 7.3890560989306504
;; LIMIT can calculate the proper answer when evaluation
;; of the function at the limit point does not:
(tan (atan +inf.0))                             ==> 16.331778728383844e15
(limit tan (atan +inf.0) -1.0e-15)              ==> +inf.0
(tan (atan +inf.0))                             ==> 16.331778728383844e15
(limit tan (atan +inf.0) 1.0e-15)               ==> -inf.0
((lambda (x) (expt (exp (/ -1 x)) x)) 0)        ==> 1.0
(limit (lambda (x) (expt (exp (/ -1 x)) x)) 0 1.0e-9)
                                                ==> 0.0
@end example