AnandTech Forums HW help in calc: SOLVED
 Register FAQ Calendar Mark Forums Read

Forums
 · Hardware and Technology · CPUs and Overclocking · Motherboards · Video Cards and Graphics · AMD Video Cards · Nvidia · Displays · Memory and Storage · Power Supplies · Cases & Cooling · SFF, Notebooks, Pre-Built/Barebones PCs · Networking · Peripherals · General Hardware · Highly Technical · Computer Help · Home Theater PCs · Consumer Electronics · Digital and Video Cameras · Mobile Devices & Gadgets · Audio/Video & Home Theater · Software · Software for Windows · All Things Apple · *nix Software · Operating Systems · Programming · PC Gaming · Console Gaming · Distributed Computing · Security · Social · Off Topic · Politics and News · Discussion Club · Love and Relationships · The Garage · Health and Fitness · Home and Garden · Merchandise and Shopping · For Sale/Trade · Hot Deals with Free Stuff/Contests · Black Friday 2015 · Forum Issues · Technical Forum Issues · Personal Forum Issues · Suggestion Box · Moderator Resources · Moderator Discussions
 10-16-2001, 02:01 AM #1 optimistic Diamond Member   Join Date: Apr 2001 Posts: 3,006 HW help in calc 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. hi thanks for your interest
 10-16-2001, 02:06 AM #2 notfred Lifer   Join Date: Feb 2001 Posts: 38,243 \$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"}
 10-16-2001, 02:06 AM #3 optimistic Diamond Member   Join Date: Apr 2001 Posts: 3,006 When adding to infite limits. Fcn's alone are defined as infinite or DNE. When adding to infite you get infite. Duh! RYan hellO?
 10-16-2001, 02:08 AM #4 Capn Platinum Member   Join Date: Jun 2000 Posts: 2,716 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
 10-16-2001, 02:22 AM #5 optimistic Diamond Member   Join Date: Apr 2001 Posts: 3,006 thanks guys! you guys care. im just dumb. this is easy adding infinite limits..........ahhhrrrgghhhhhh..maybe i need sleep