# Thread: Trig Limit when the Numerator is Undefined

1. ## Trig Limit when the Numerator is Undefined

Can somebody help me understand this one, please?

$\displaystyle \lim_{x\to \infty } \frac{\cos^4(x)}{5 + x^2}$

I get that the answer is zero, because the denominator goes to infinity due to the $x^2$. I also understand that the value of $\cos^4(x)$ is always between 0 and 1, which is why we can conclude that the limit overall is zero.

What is a little weird for me is that the limit of $\cos^4(x)$ doesn't exist. I can't get past the idea that the limit of a fraction can be defined when the limit of the numerator doesn't exist. What if the numerator was simply $\cos (x)$? Then what would the overall limit be (since $\cos (x)$ can vary between -1 and 1)?

Sorry if I'm being a little dense, but can somebody try to explain this to me?

Thanks.

2. Originally Posted by joatmon
Can somebody help me understand this one, please?

$\displaystyle \lim_{h\to \infty } \frac{\cos^4(x)}{5 + x^2}$

I get that the answer is zero, because the numerator goes to infinity due to the $x^2$. I also understand that the value of $\cos^4(x)$ is always between 0 and 1, which is why we can conclude that the limit overall is zero.

What is a little weird for me is that the limit of $\cos^4(x)$ doesn't exist. I can't get past the idea that the limit of a fraction can be defined when the limit of the numerator doesn't exist. What if the numerator was simply $\cos (x)$? Then what would the overall limit be (since $\cos (x)$ can vary between -1 and 1)?

Sorry if I'm being a little dense, but can somebody try to explain this to me?

Thanks.
The idea is hidden in what you have said. The squeeze theorem gives you what you need. You have said that you believe that
$\displaystyle -1 \le \cos(x) \le 1$ for all x. Now if we divide this inequality by $x^2+1$ Since this is always non negative we don't have to change the inequality signs we get

$\displaystyle \frac{-1}{x^2+1} \le \frac{\cos(x)}{x^2+1} \le \frac{1}{x^2+1}$

This inequality is still valid for all values of x

Now if we take the limit with out knowing what the middle limit its we will know two numbers that it is still in between by the inequalities.

$\displaystyle \lim_{x \to \infty}\frac{-1}{x^2+1} \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le \lim_{x \to \infty}\frac{1}{x^2+1}$

This gives

$\displaystyle 0 \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le 0$

This tells us that the limit is bigger than $0$, but also less than $0$ so the limit must be zero.

3. I assume this should actually be $\displaystyle \lim_{x \to \infty}\frac{\cos^4{x}}{5 + x^2}$

I think you can squeeze it...

$\displaystyle \frac{0}{5 + x^2} \leq \frac{\cos^4{x}}{5 + x^2} \leq \frac{1}{5 + x^2}$, so

$\displaystyle \lim_{x \to \infty} 0 \leq \lim_{x \to \infty}\frac{\cos^4{x}}{5 + x^2} \leq \lim_{x \to \infty}\frac{1}{5 + x^2}$

$\displaystyle 0 \leq \lim_{x \to \infty}\frac{\cos^4{x}}{5 + x^2} \leq 0$.

So $\displaystyle \lim_{x \to \infty} \frac{\cos^4{x}}{5 + x^2} = 0$.

4. Thanks. I had played with trying to squeeze this out, but couldn't get it to work.

So, my logic about why the limit is zero isn't correct. I had thought that, since the function $cos^4(x)$ is always between 0 and 1, which we know even without using the squeeze theorem, what forces the limit to zero is that the denominator becomes infinitely large.

But what I think that you are saying is that the denominator has nothing to do with it. It's that $\lim_{x\to \infty } \cos^4(x)$ can be shown to be zero by using the squeeze theorem. Thus, no matter what happens to the denominator (so long as its limit exists and is not zero), the value of the limit is still zero since the limit of the numerator is zero.

Is a correct interpretation?

5. Just saw Prove It's reply. Now I get it entirely. Thanks to both of you.

6. Originally Posted by TheEmptySet
The idea is hidden in what you have said. The squeeze theorem gives you what you need. You have said that you believe that
$\displaystyle -1 \le \cos(x) \le 1$ for all x. Now if we divide this inequality by $x^2+1$ Since this is always non negative we don't have to change the inequality signs we get

$\displaystyle \frac{-1}{x^2+1} \le \frac{\cos(x)}{x^2+1} \le \frac{1}{x^2+1}$

This inequality is still valid for all values of x

Now if we take the limit with out knowing what the middle limit its we will know two numbers that it is still in between by the inequalities.

$\displaystyle \lim_{x \to \infty}\frac{-1}{x^2+1} \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le \lim_{x \to \infty}\frac{1}{x^2+1}$

This gives

$\displaystyle 0 \le \lim_{x \to \infty}\frac{\cos(x)}{x^2+1} \le 0$

This tells us that the limit is bigger than $0$, but also less than $0$ so the limit must be zero.
Having the fourth power of cosine, I'd use 0 as the lower bounds, since the numerator and denomiator are positive.
I see that Profit said the same thing.

7. Originally Posted by joatmon
Thanks. I had played with trying to squeeze this out, but couldn't get it to work.

So, my logic about why the limit is zero isn't correct. I had thought that, since the function $cos^4(x)$ is always between 0 and 1, which we know even without using the squeeze theorem, what forces the limit to zero is that the denominator becomes infinitely large.

But what I think that you are saying is that the denominator has nothing to do with it. It's that $\lim_{x\to \infty } \cos^4(x)$ can be shown to be zero by using the squeeze theorem.
No, it doesn't. The constant function f(x)= 1/2 is always between 0 and 1- does that "force" its limit to be 0?
$\lim_{x\to\infty} cos^4(x)$ does not exist.

Thus, no matter what happens to the denominator (so long as its limit exists and is not zero), the value of the limit is still zero since the limit of the numerator is zero.

Is a correct interpretation?
No, it isn't. It is the fact that the denomator goes to infinity that makes the limit 0. The only thing required of the numerator is that it NOT go to infinity or negative infinity.