# Tough limits

1. Apr 29, 2014

### negation

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

The question states:

It is given limit f(x) as x->0 is 7
what should we then expect of the limit f(1/x) as x->∞

3. The attempt at a solution

0? I'm handwaving an answer because I have no idea the approach towards this problem. One states the limit x->0 and the other x->∞. I'm not seeing a connection.

2. Apr 29, 2014

### PAllen

What is the limit of 1/x as x->∞?

Have you learned any rules about limits over compounding of functions, e.g. f(g(x)) ?

3. Apr 29, 2014

### negation

The limit is 0 as x tends towards infinity.

Yes but how does it relates to this problem?

4. Apr 29, 2014

### PAllen

Think. I'm afraid I can't give another hint without just giving the answer.

5. Apr 29, 2014

### negation

I'm only posting because its been a couple of hours thinking. I really have no idea as to the relevance.

6. Apr 29, 2014

### SammyS

Staff Emeritus
Think about PAllen's hint. ... then think about your response to that hint. ... then how does that fit back into the given problem?

then ... repeat.

Think about PAllen's hint. ... then think about your response to that hint. ... then how does that fit back into the given problem?

then ... repeat.

...

7. Apr 29, 2014

### Mentallic

The function f(x)=x+7 has the property that when x approaches 0, f(x)=7. The limit of this function at x=0 is 7.
f is the operation we do to our variable, in this case, x, and that operation in this case is to add 7 to our variable.

If we changed our variable to anything else, say, a, then we have
f(a) = a+7

And nothing here has changed except for the symbol we used for our variable. The limit as a goes to 0 will still be 7.

Now if we let a = 1/x, we have

f(1/x) = 1/x + 7

And if x approaches infinity, what is our limit?

8. Apr 29, 2014

### negation

I arrived at 7 which is consistent with the answer key I have ( This shows the answer means nothing to me if I cannot understand).

lim f(x) as x-> 0 = 7 where f(x) = x+7
(is it necessary that we add x as a variable? It is true also that without the x variable, the limit is 7.)

The limit f(1/x) = lim x->∞ = 7+1/x = 7+0 = 7

Last edited: Apr 29, 2014
9. Apr 29, 2014

### PAllen

if you know about limits for compounding functions, let g(x)=1/x, and apply a rule you know.

10. Apr 29, 2014

### Mentallic

It's really not that difficult.

$$\lim_{u\to 0} f(u) = 7$$

If we ignore that f(u) is likely undefined at u=0, then this is pretty much like evaluating f(0).

$$\lim_{x\to \infty}f(1/x)$$

Is also pretty much like evaluating f(0) because

$$\lim_{x\to \infty} 1/x = 0$$

Hence the two limits are asking the exact same question.

The x variable isn't necessary, but I wanted to use a simple function for the example without possibly causing more confusion.

EDIT:

Maybe it'll help if you saw a few more examples.

$$\lim_{x\to 1} g(x) = 3/2$$
So in essence, g(1) = 3/2.

Then it's also true that
$$\lim_{x\to 1} g(1/x) = 3/2$$
Because 1/x = 1/1 = 1, hence g(1/x) = g(1) = 3/2

Also,

$$\lim_{x\to 1} g(x^n) = 3/2$$
For any n since if x=1, 1n = 1 and so g(1n)=g(1)=3/2.

But,

$$\lim_{x\to 1} g(x+1)$$

is likely not equal to 3/2 because we are now evaluating g(2) and that is probably a different value to g(1).

Last edited: Apr 29, 2014
11. Apr 29, 2014

### PAllen

For a proof, you have to be careful of the assumptions for compounding limits. Suppose you believe the following:

given lim y->0 f(y) = 1
and limit x->0 g(x) = 0

then limit x->0 f(g(x)) = 1

This could be false. Consider that f(y) = 1 for y ≠ 0, f(y)=0 for y=0. Consider that g(x)=0. Then the above 'theorem' would claim lim x->0 f(g(x)) = 1, while it is really 0.

The original problem statement did not say f is continuous, so this type of case is not excluded. You want to look at compounding theorems that cover the possibility of f not being continuous. A key assumption for that case is that if g(x) ≠ 0 is true (while lim x->0 g(x) = 0 is still true), then with this extra assumption, the theorem would be true. In the given problem here, 1/x is never 0, so the compounding rule is true even if f is not continuous at its limit point.