# L'Hôpital's rule and weird expression

1. Oct 27, 2013

### fogvajarash

1. The problem statement, all variables and given/known data
Find, using l'Hôpital's rule:

lim x-> ∞ ((tan(ax)-atanx)/(sin(ax)-asinx))

Where a is a non constant greater than ± 1.

2. Relevant equations
-

3. The attempt at a solution
I can't work out anything from this point. I don't know how to change or factorize the expression (is there a possible way to separate the (ax) angle?

Last edited: Oct 27, 2013
2. Oct 27, 2013

### Ray Vickson

Try plotting some numerical examples for a range of large x (say with a = 1.5 and similar values).

3. Oct 27, 2013

### Staff: Mentor

What does this mean? Since the limit is as x varies, a is a constant in this process. Also, by "greater than ± 1", I presume you mean a > 1 or a < -1, or more briefly, |a| > 1.

4. Oct 27, 2013

### Ray Vickson

You really should do what I suggested in my first response; it will help you to realize there is something seriously wrong with the question as written. I will not say any more until I see a response from you.

5. Oct 27, 2013

### brmath

yep (they don't like if I submit a one word answer , but yep is what I think)

6. Oct 27, 2013

### fogvajarash

I've done it and it seems that the function does not tend to any value. How can then we argue that the limit does not exist? (if it tends to 0, then we can have a limit as it is an indeterminate form). However when it tends to infinity i'm not sure on how to argue on that.

7. Oct 27, 2013

### Ray Vickson

That is what I meant when I claimed there is something seriously wrong with the statement of the problem: there is no limit and so the problem makes no sense. Do you think maybe they wanted the limit as x → 0 rather than x → ∞? The zero-limit problem does have meaning!

8. Oct 27, 2013

### fogvajarash

I think they actually asked for x → 0, but I can't be sure on that (a friend just showed me the problem from his class).

Is there a formal way to state that the limit of a function does not exist? (in the case of sine and cosine we can say that it doesn't exist as it doesn't approach any value, or when the RHS and LHS limits are not equal. But can we do something in this case?)

9. Oct 27, 2013

### Ray Vickson

Yes: a limit does not exist if there is no finite number, L, to which the values of f(x) tend as x → ∞. You can see that a limit does not exist in this case because we have $f(r_n) > 10$ for some infinite sequence $r_n \to +\infty$ and $f(s_n) < -10$ for some other infinite sequence $s_n \to +\infty$. The graph shows that there are infinitely many values where f tends to +∞ and infinitely many values where f tends to -∞; certainly, then, there are infinitely many values where f > 10 and others where f < -10.

I suppose one should not really trust a graph, but should show analytically that these statements hold true. All you need show is that there are infinitely many values where the denominator goes to 0 but the numerator does not. (Also, there are points where the numerator tends to infinity and the denominator remains bounded, but that just strengthens the case.)

10. Oct 27, 2013

### brmath

The formal way to say the limit of a function does not exist is to say "the limit of this function does not exist at this point".

11. Oct 27, 2013

### Dick

If the limit is x->0, and I really think that's what they meant, then l'Hopital's will work. You'll need multiple applications to get the result though.

Last edited: Oct 28, 2013
12. Oct 28, 2013

### BruceW

yeah, that's an interesting question. Mathematically, for a limit to exist, we are saying that:
$$\lim_{x\rightarrow \infty} f(x) = L$$
And what this means is: for all $\varepsilon >0$, there exists $c$ such that $|f(x)-L|<\varepsilon$ whenever $x>c$. (This is the epsilon-delta definition of limit, and is the standard one I think). So, to prove that the limit does not exist, you would have to prove that the above statement is not true for your function. (for $L$ being any real number).