Does the Existence of lim f(x)g(x) Imply the Existence of lim f(x) and lim g(x)?

[babita]

Homework Statement

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

Homework Equations

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
pleasezzzzzzzzzzzzzzzzzzzzzzzzzzz help

 

[Mark44]

Homework Statement

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

Homework Equations

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
pleasezzzzzzzzzzzzzzzzzzzzzzzzzzz help

$$\lim_{x \to 0} f(x)g(x) = \lim_{x \to 0}f(x)\cdot \lim_{x \to 0} g(x)$$
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.

 

[babita]

got it, thanks:)
please 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 -∞
please explain...

 

[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 shouldn't the answer be 1? the book says -1

 

[Mark44]

got it, thanks:)
please 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 -∞

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
$\lim_{x \to 0^+}\frac{1}{x} = \infty$
$\lim_{x \to 0}\frac{-1}{x^2} = -\infty$

 

[babita]

And how I'm wrong here:
xsin(1/x)here if x tends to -∞ it would be sin(-1/ ∞)/-1/ ∞
so shouldn't the answer be 1?

 

[Mark44]

And how I'm wrong here:
xsin(1/x)here if x tends to -∞ it would be sin(-1/ ∞)/-1/ ∞
so shouldn't the answer be 1?

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

 

[HallsofIvy]

And how I'm wrong here:
xsin(1/x)here if x tends to -∞ it would be sin(-1/ ∞)/-1/ ∞
so shouldn't the answer be 1?

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.

 

[babita]

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