# HW help in calc: SOLVED

Discussion in 'Off Topic' started by optimistic, Oct 16, 2001.

1. ### optimistic Diamond Member

Joined:
Apr 29, 2001
Messages:
3,006
0
Show by means of an example that lim x->a [f(X) + g(x)] may exist even though neither lim x->a f(x) nor line x->a g(X) exists.

#1
2. ### notfred Lifer

Joined:
Feb 12, 2001
Messages:
38,243
0
\$f = "lim x->a f(x)";
\$g = "line x->a g(X)";
\$lim = "x->a [f(X) + g(x)]";

\$f =~ s/[ -}]//g;
\$g =~ s/[ -}]//g;

if (\$f ne ''){print "limit exists\n"}
else{print "limit doesn't exist\n"}

if (\$g ne ''){print "limit exists\n"}
else{print "limit doesn't exist\n"}

if (\$lim ne ''){print "limit exists\n"}
else{print "limit doesn't exist\n"}

#2
3. ### optimistic Diamond Member

Joined:
Apr 29, 2001
Messages:
3,006
0
When adding to infite limits. Fcn's alone are defined as infinite or DNE. When adding to infite you get infite.

Duh! RYan hellO?

#3
4. ### Capn Platinum Member

Joined:
Jun 27, 2000
Messages:
2,716
0
exist as in a finite number?

If that's the case then

f(x) = 1/(2-x)

g(x) = 1/(x-2)

x->2 for f(x) would be +/- infinity depending on direction
x->2 for g(x) would also be +/- infinity depending on direction

however f(x)+g(x) equals 0, so x->2 equals 0

Not sure if that's what you're looking for

#4

Joined:
Apr 29, 2001
Messages:
3,006