# Limit problem

1. Jan 31, 2015

### Dumbledore211

1. The problem statement, all variables and given/known data
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
2. Relevant equations
No relevant equations are required in this problem. To determine whether the statement is true or false

3. 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.

2. Jan 31, 2015

### BvU

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: Jan 31, 2015
3. Jan 31, 2015

### Staff: Mentor

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.
I agree that the first post was unclear, and that the OP meant division instead of multiplication.

4. Jan 31, 2015

### Ray Vickson

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 $x \downarrow a$.

5. Jan 31, 2015

### 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.