# Limits of f(x) & g(x) Do Not Equal Lim[f(x)g(x)]

• Dumbledore211
I sometimes get carried away when I'm learning things but I think I've corrected the OP's confusion.##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ##It's false, and the reason is that you're dividing by 0. In the particular case of ##f(x) = x-a, g(x) = (x-a)^2## you get ##x \to a\;:\;\;f(x)/g(x) = 1/(x-a)## which doesn't have a limit at a and the limit of the product is 0, not 1.In summary, the statement that if ##\displaystyle \lim_{x \to a

## Homework Statement

If Lim f(x) and Lim g(x) both exist and are equal
x→a x→a then Lim[f(x)g(x)]=1
x→a

## Homework Equations

No relevant equations are required in this problem. To determine whether the statement is true or false [/B]

## The Attempt at a Solution

The statement is false but the reason behind it is quite unclear to me. Is it because when the limit approaches a from left and right the limits are -∞ and +∞?? Would be very helpful if anyone of you could explain it.

If ##\displaystyle \lim_{x\rightarrow a} f(x)## and ##\displaystyle \lim_{x\rightarrow a} g(x)## both exist and are equal ##\Rightarrow \displaystyle \lim_{x\rightarrow a} f(x)g(x) = 1 ## ?

Could that be ##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ## ?

• To me ##\displaystyle \lim_{x\rightarrow a}## means "x approaching from the left to a" so I wouldn't worry about "coming from the right"
• To me ##\displaystyle \lim_{x\rightarrow a} f(x) = \pm \infty## means the limit does not exist, so I wouldn't worry about those either
• But the division cause a problem in a particular case, that can serve as a counter-example that makes the general statement not true,

Last edited:
Ray Vickson
BvU said:

If ##\displaystyle \lim_{x\rightarrow a} f(x)## and ##\displaystyle \lim_{x\rightarrow a} g(x)## both exist and are equal ##\Rightarrow \displaystyle \lim_{x\rightarrow a} f(x)g(x) = 1 ## ?

Could that be ##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ## ?

• To me ##\displaystyle \lim_{x\rightarrow a}## means "x approaching from the left to a" so I wouldn't worry about "coming from the right"
The notation ##\lim_{x \to a} f(x)## indicates a two-sided limit in which x can approach a from either side. If the two-sided limit exists, then both one-sided limits also exist. Maybe that's what you were trying to say, but what you actually said wasn't clear.
BvU said:
• To me ##\displaystyle \lim_{x\rightarrow a} fx) = \pm \infty## means the limit does not exist, so I wouldn't worry about those either
• But the division cause a problem in a particular case, that can serve as a counter-example that makes the general statement not true,
I agree that the first post was unclear, and that the OP meant division instead of multiplication.

BvU
BvU said:

If ##\displaystyle \lim_{x\rightarrow a} f(x)## and ##\displaystyle \lim_{x\rightarrow a} g(x)## both exist and are equal ##\Rightarrow \displaystyle \lim_{x\rightarrow a} f(x)g(x) = 1 ## ?

Could that be ##\displaystyle \lim_{x\rightarrow a} f(x)/g(x) = 1 ## ?

• To me ##\displaystyle \lim_{x\rightarrow a}## means "x approaching from the left to a" so I wouldn't worry about "coming from the right"
• To me ##\displaystyle \lim_{x\rightarrow a} fx) = \pm \infty## means the limit does not exist, so I wouldn't worry about those either
• But the division cause a problem in a particular case, that can serve as a counter-example that makes the general statement not true,

Standard usage is that ##x \to a## means a two-sided limit (##|x-a| \to 0##). Left-hand limits are typically denoted as ##x \to a\!-## or ##x \to a\!-\!0## or ##x \uparrow a##. Right-hand limits are typically denoted as ##x \to a\!+## or ##x \to a\!+\!0## or ##x \downarrow a##.

BvU
Good thing you corrected me, gentlemen. Been too long ago, but I do recognize the ##x\downarrow a## and ##x\uparrow a\;.\ ## Never had much opportunity to make good use of the distinction. Apologies to professor D.