# On properties of limits

1. Mar 27, 2012

### babita

1. The problem statement, all variables and given/known data

the ques says:
lim x tending to 0 [f(x)g(x)] exists. Then both lim x tending to 0 f(x) AND lim x tending to 0 also exist. True or False

2. Relevant equations

3. The attempt at a solution
lim f(x)g(x) =lim f(x) * lim g(x)
so if LHS exists then limf(x) and lim g(x) must exist
so it should be true
but the answer says that the statement is false
plzzzzzzzzzzzzzzzzzzzzzzzzzzzz help

2. Mar 27, 2012

### Staff: Mentor

$$\lim_{x \to 0} f(x)g(x) = \lim_{x \to 0}f(x)\cdot \lim_{x \to 0} g(x)$$
only if both limits on the right side exist.

What you need is a counterexample in which the limits on the right side fail to exist, but the limit of the product does exist.

3. Mar 27, 2012

### babita

got it, thanks:)
plz help me out on this one also:
i was doing limits and i noticed that while 1/ ∞ and -1/ ∞ both are taken as 0
the author writes 1/0 as + ∞ and -1/0 as -∞

4. Mar 27, 2012

### babita

i understand that -0 and +0 would mean same and -∞ and +∞ mean different
but i'm getting some answers wrong. (and i'm getting confused)
for eg: xsin(1/x)here if x tends to -∞ it would be sin(-1/ ∞)/-1/ ∞
so shouldnt the answer be 1? the book says -1

5. Mar 27, 2012

### Staff: Mentor

You'll need to show me what the author is actually saying. Both 1/0 and -1/0 are undefined, and a limit that has one of these forms can turn out to be ∞, -∞, or fail to exist completely.

Some examples:
$\lim_{x \to 0}\frac{1}{x}$ does not exist
$\lim_{x \to 0^+}\frac{1}{x} = \infty$
$\lim_{x \to 0}\frac{-1}{x^2} = -\infty$

6. Mar 27, 2012

### babita

And how i'm wrong here:
xsin(1/x)here if x tends to -∞ it would be sin(-1/ ∞)/-1/ ∞
so shouldnt the answer be 1?

7. Mar 27, 2012

### Staff: Mentor

Yes, but you should never write expressions such as sin(-1/∞) and the like. That's what limits are for.

8. Mar 28, 2012

### HallsofIvy

Staff Emeritus
As Mark44 said, you should never write things like "1/0", $1/\infty$, or $1/-\infty$. To find the limit of x sin(1/x) as x goes to $-\infty$, let y= 1/x so we have $\lim_{y\to 0^-} sin(y)/y$. That can be easily shown (how depending on exactly how you have defined sine) to be 1.

9. Mar 28, 2012

### babita

i understand that......just wrote it to explain what i was thinking
thanks