HI,
Here is a program for HP49/50 whch try to find the lim (X,Y)->(a,b) of a function of two variables.
Just put your function on the stack (variables must be X,Y) , and the { a b } limit you search. You will see in return :
lim:number is the limit found by the program
lim: oo (infinite limit)
lim:? no limit found
lim: f(t) no limit, the function returned is the limit in function of t, the angle of approach.
Must be in RPN and exact mode. Be careful about X Y r t not define as global variables. If you try it, you will see some limitations but in most cases it works fine.«
{'X' 'Y'} DUP ROT ADD = + « SUBST » STREAM
DUP
'X=0' SUBST
'Y=0' SUBST
IF EVAL DUP '?' SAME THEN
DROP
'X=r*COS(t)' SUBST
'Y=r*SIN(t)' SUBST
SIMPLIFY
IFERR 'r=0+0' lim THEN DROP2 '?' END
ELSE
NIP
END
SIMPLIFY 'Lim' ->TAG
»
'LimXY' STOFor example :
'3*X^2*Y/(X^2+Y^2)' { 0 0 } LimXY. Here you get :
lim:0'(X^2-Y^2)/(X^2+Y^2)' {0 0}
Lim: '-(2*SIN(t)^2-1) that means no limit (as the limit depands of the
angle to approach the point). You can modifiy this if you prefer with '?'
and no detail'X*SIN((X+Y)/4)' { 'PI' 'PI' }
Lim:PI'LOG(X+e^Y)/Sqrt(X^2+Y^2)' { 1 0 }
Lim:LOG(2)'LN(X+Y)/(X^2+2*X*Y+Y^2-1)' { 1 0 }
Lim:1/2'(X^2*Y^5)/(X^4+X^2*Y^2+Y^4)' {0 0}
Lim : 0'8*(X^2*Y^2)/(X^4+Y^4)' {0 0}
Lim: '-((8*SIN(t)^4-8*SIN(t)^2)/(2*SIN(t)^4-2*SIN(t)^2+1))' @ Mean the
limit does not existSIN(X+Y)/(X+Y) { 0 0 }
Lim: 1' (X^2+Y^2-2*X-2*Y)/(X^2+Y^2-2*X+2*Y+2)' { 1 -1 }
Lim: oo (infinty)' (X^3+X*Y^2)/(X^2+Y^2)' { 0 0 }
Lim : 0etc.
Edited: 26 Aug 2012, 10:34 a.m.