iscont - Maple Help

iscont

test continuity on an interval

 Calling Sequence iscont(expr, x = a .. b) iscont(expr, x = a .. b, 'closed') iscont(expr, x = a .. b, 'open')

Parameters

 expr - an algebraic expression x - variable name a..b - real interval 'closed' - (optional) indicates that endpoints should be checked 'open' - (optional) indicates that endpoints should not be checked (default)

Description

 • The iscont function returns true if the expression is continuous on the interval, or false if the expression is not continuous.  If iscont cannot determine the result it returns FAIL.
 • The iscont function assumes that any symbols in the expression are real.
 • By default, or if the optional third argument 'open' is specified, then the interval is considered to be an open interval. If the optional third argument 'closed' is specified then continuity at the endpoints is checked;  specifically under this option, $\underset{x→{a}^{+}}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{expr}$ and $\underset{x→{b}^{-}}{lim}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\mathrm{expr}$ must be defined and be finite.
 • The endpoints of the interval must be either real constants or infinity or -infinity.
 • If the endpoints of the interval are ordered with the first value greater than the second, iscont will reverse them.

Examples

 > $\mathrm{iscont}\left(\mathrm{tan}\left(x\right),x=0..1\right)$
 ${\mathrm{true}}$ (1)
 > $\mathrm{iscont}\left(\mathrm{tan}\left(x\right),x=0..2\mathrm{π}\right)$
 ${\mathrm{false}}$ (2)
 > $\mathrm{iscont}\left(\mathrm{piecewise}\left(x<3,x+8,3\le x,{x}^{2}+2\right),x=0..\mathrm{∞}\right)$
 ${\mathrm{true}}$ (3)
 > $\mathrm{iscont}\left(\mathrm{piecewise}\left(x<3,x+2,3\le x,{x}^{2}+2\right),x=0..\mathrm{∞}\right)$
 ${\mathrm{false}}$ (4)
 > $\mathrm{iscont}\left(\frac{1}{x},x=1..2\right)$
 ${\mathrm{true}}$ (5)
 > $\mathrm{iscont}\left(\frac{1}{x},x=-1..1\right)$
 ${\mathrm{false}}$ (6)
 > $\mathrm{iscont}\left(\frac{1}{x},x=-\mathrm{∞}..\mathrm{∞}\right)$
 ${\mathrm{false}}$ (7)

Unless specified, the interval is assumed to be open.

 > $\mathrm{iscont}\left(\frac{1}{x},x=0..1\right)$
 ${\mathrm{true}}$ (8)
 > $\mathrm{iscont}\left(\frac{1}{x},x=0..1,'\mathrm{closed}'\right)$
 ${\mathrm{false}}$ (9)

If iscont cannot determine the result it returns FAIL.

 > $\mathrm{iscont}\left(\frac{1}{x+a},x=0..1\right)$
 ${\mathrm{FAIL}}$ (10)
 > $\mathrm{iscont}\left(\frac{1}{x},x=3..1\right)$
 ${\mathrm{true}}$ (11)