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SYMBMATH.H36
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4.4 Limits
4.4.1 One-sided Limits
You can finds real or complex limits, and discontinuity or
one-sided value.
First find the expression value by subs(y, x = x0) or the
function value by f(x0) when x = x0.
If the result is the discont (i.e. discontinuity), then use the
one-sided value x0+zero or x0-zero to try to find the one-sided function
or expression value.
For a function f(x), you can evaluate the left- or right-sided
function value, similar you evaluate the normal function value:
f(x0-zero)
f(x0+zero)
For an expression y, you can evaluate its one-sided expression
value by
subs(y, x = x0-zero)
subs(y, x = x0+zero)
The discont (discontinuity) means that the expression has
a discontinuity and only has the one-sided value at x=x0. You should
use x0+zero or x0-zero to find the one-sided value. The value of
f(x0+zero) or f(x0-zero) is the right-sided or left-sided function value
as approaching x0 from positive (+inf) or negative (-inf) direction,
respectively, i.e. as x = x0+ or x = x0-.
If the result is undefined (indeterminate forms, e.g. 0/0,
inf/inf, 0*inf, and 0^0), then find its limit by
lim(y, x = x0)
If the limit is discont, then you can find a left-
or right-sided limit when x approaches to x0 from positive (+inf)
or negative (-inf) direction at discontinuity by
lim(y, x = x0+zero)
lim(y, x = x0-zero)
Example 4.4.2.
Evaluate y=exp(1/x) at x=0, if the result is discontinuty, find
its left-sided and right-sided values (i.e. when x approaches 0 from
positive and negative directions).
IN: y:=exp(1/x)
IN: subs(y, x = 0)
OUT: discont # discontinuty at x=0
IN: subs(y, x = 0+zero), subs(y, x = 0-zero)
OUT: inf, 0
Example 4.4.3:
How to handle the following one-sided values ?
Let f(x) = 1 when x < 1, f(x) = 1 when x > 1 (and not defined at x = 1).
Let g(x) = 1 when x < 1, g(x) = 1 when x > 1, and g(1) = 2.
Let h(x) = 1 when x < 1, h(x) = 2 when x >= 1.
Let k(x) = 1 when x < 1, k(x) = 2 when x > 1, and k(1) = 3.
Now ask SymbMath to compute
(1) the limit as x approaches 1,
(2) the limit as x approaches 1 from the left, and
(3) the limit as x approaches 1 from the right
for each of the above piecewise defined functiuons.
# define functions
f(x_) := if(x<1 or x>1, 1)
f(1+zero):=1
f(1-zero):=1
g(x_) := if( x<1 or x>1, 1)
g(1):=2
g(1+zero):=1
g(1-zero):=1
h(x_) := if( x<1, 1, 2)
h(1+zero):=2
h(1-zero):=1
k(x_) := if( x<1, 1, if( x>1, 2))
k(1):=3
k(1+zero):=2
k(1-zero):=1
# evaluate functions
IN: f(1), g(1), h(1), k(1)
OUT: f(1), 2, 2, 3
IN: f(1+zero), g(1+zero), h(1+zero), k(1+zero)
# right-hand side value at x=1+
OUT: 1, 1, 1, 1
IN: f(1-zero), g(1-zero), h(1-zero), k(1-zero)
# left-hand side value at x=1-
OUT: 1, 1, 2, 2
Example 4.4.1.
Find limits of types 0/0 and inf/inf.
IN: p:=(x^2-4)/(2*x-4)
IN: subs(p, x = 2)
OUT: undefined
IN: lim(p, x = 2)
OUT: 2
IN: subs(p, x = inf)
OUT: undefined
IN: lim(p, x = inf)
OUT: inf
4.4.2 Numeric limits: NLim()
If symbolic limit sometines fall, you should try numeric limit by
nlim(y, x=x0)