# Limit as x approaches infinity

1. Nov 4, 2008

### aquitaine

1. Find the limit as x approaches infinity of (cos(1/x))^(x^2)

attempt at a solution
I tried using e to change it's form to e^(ln(cos(1/x)*x^2) and taking the limit of the power, the problem is I'm really stuck at this point with the limit as x approaches infinity of ln(cos(1/x))*x^2, since l'hospital's rule won't work. Any suggestions for what to do next?

2. Nov 4, 2008

### gabbagabbahey

Try a substitution of the form $u=\frac{1}{x}$...that should allow you to use l'Hopital's rule.

3. Nov 4, 2008

Thanks.

4. Nov 5, 2008

### mrjoe2

what is the point of that. just substitute dude!

lim
x->inf cos(1/inf) ^inf^2
= cos(0) ^inf^2
= 1 ^inf^2
= 1

5. Nov 5, 2008

### gabbagabbahey

The limit is not one!....try plugging in some large values of x and see what number you approach..

6. Nov 5, 2008

### mrjoe2

the limit is still 1 i believe. there is no flaw in my work bro.

7. Nov 5, 2008

### mrjoe2

i just subbed it into my graphing calculator and it is one. it makes sense logically too, as well as through limit laws and common sense

8. Nov 5, 2008

### mrjoe2

also im not sure its plausible to use l'hopitals rule for limits of x approaching infinite*. that is an important note that most students i tutor forget. they think it can be used for everything, when it is wrong ALOT of the time. i only use it as a last resort

9. Nov 5, 2008

### gabbagabbahey

The value of a function at x=infinity is not always the same as the limit as x approaches infinity of the function; this is the case here. The problem is that cos(1/x))^(x^2) is not a rational function.

The formal definition of the limit at infinity for such a function is:

Is that condition satisfied for this function with $L=1$?....I think not.

10. Nov 5, 2008

### HallsofIvy

Staff Emeritus
No, it's not.

Is your calculator in 'degree' or 'radian' mode? It has to be in radian mode for this problem.

That's why gabbagabbahey's hint was to substitute u= 1/x and THEN use L'Hopital's Rule. With u= 1/x, $\lim_{x\rightarrow \infty} (cos(1/x))^{x^2}$ becomes $\lim_{u=\rightarrow 0}(cos(u))^{1/u^2}$. Taking the logarithm of that we have $(1/u) ln(cos(u))= ln(cos(u))/u^2$ which is 0/0, the standard "L'Hopital's Rule" case.

Use LHopital's rule to find the limit of ln(cos(u))/u2. The limit of the original function will be e to that power.

This is twice now someone, in recent days, has protested that L'Hopital's rule could not be used when, in fact, the problem could be set into a form in which it could be used. The other was where a person had a limit of a sequence an and it was suggested that they use L'Hopitals' rule. Someone, I don't recall if it was mrjoe2, protested that L'Hopital's rule could only be used with functions, not, sequences. I pointed out that, if there exist a function f(x) such that an= f(n) for all n, then if f(x) converges to limit L, as x goes to infinity, so must any sequence f(xn) where an goes to infinity and certainly f(n) fits that. If you are tutoring Calculus and believe that your post #4 was correct, you might want to review sequences.

11. Nov 5, 2008

### Skatch

mrjoe, your substitution method is by no means rigorous. In simple cases, it can give you the correct answer, but the limit of this problem is definitely not 1.

The original poster was correct to change it to a power of e, then study the limit of the exponent. You get e raised to the power:

ln(cos(1/x))^x2

which equals:

x2*ln(cos(1/x)).

You can rewrite this as:

ln (cos(1/x)) / (1/x2)

And then use L'Hopital's rule (because you have 0/0 as x->Inf.)

You can show the work to see that this equals:

tan(1/x) / (-2/x)

Use L'Hopital's again and you will end up with a fraction. Your limit then is e raised to this fraction.

Last edited: Nov 5, 2008