Understanding Non-Existing Function Limits: Quick Help Needed

[matrix_204]

lets say there r two functions, f( x) and g(x ), and the limits of both of these function as x approaches c, does not exist but the lim(f(x )*g(x )) can still exist as x approaches c.

I have tried doing several examples of functions that don't exist and then i got confused because if the two functions f and g, don't exist then how am i suppose to kno their values, and when i multiply the two, how can the multiple of two non-existing functions exist. need help as fast as possible, thanks

 

[AKG]

f({x | x < c}) = {0}, f({x | x > c}) = {1}. Use this and a similar definition for g.

 

[matrix_204]

yess, i tried that, but i don't understand the part, where wen u multiply the two, the function exists, how do u kno the value of the non-existing functions

 

[Tide]

How about

$$f(x) = e^{\frac {1}{x-c}$$

and

$$g(x) = e^{\frac {-1}{x-c}$$

though I'd still have a problem concering whether the product is defined?

 

[matrix_204]

well i sumhow have done it using an example, but it doesn't seem to me like that it is true for every non-existing functions, but ne ways, since the question asks for a specific example, i don't think there is anything more i could do. Btw i haven't really learned the e notation practically yet, maybe next semester, lol, but thanks anyways.

 

[AKG]

Huh? I don't know why you're talking about non-existing functions. If you tried what I suggested in my post, I can't see where you would be having problems. Perhaps you did not understand, but I didn't think too much explanation was need for such a problem. Anyhow:

Define f(x) = 1 for x < c, and f(x) = 0 otherwise. Similarly, define g(x) = 0 if x < c, and g(x) = 1 otherwise. Clearly, the limits of the functions do not exist individually at c, since the value of the function jumps there (either immediately from 0 to 1, or from 1 to 0). But, multiply these two functions, and you will obviously have that g(x)*f(x) = 0 for any x, so, of course, the limit is zero everywhere.