1. ## Limit Question

$\displaystyle \mathop {\lim }\limits_{x \to \infty } \frac{{\cos x}} {{e^x }} = 0$

L'Hopitals rule really doesn't help in this situation. How would you determine this? As x approaches infinity, the limit of cosine is indefinite. Also as x approaches infinity, e^x approaches infinity.

So basically, are you supposed to look at it from the perspective that no matter what is going on with cosine, e^x becomes so massive that the numerator is negligible?....or how should you approach this limit?

Thank you.

2. Originally Posted by RedBarchetta
$\displaystyle \mathop {\lim }\limits_{x \to \infty } \frac{{\cos x}} {{e^x }} = 0$

L'Hopitals rule really doesn't help in this situation. How would you determine this? As x approaches infinity, the limit of cosine is indefinite. Also as x approaches infinity, e^x approaches infinity.

So basically, are you supposed to look at it from the perspective that no matter what is going on with cosine, e^x becomes so massive that the numerator is negligible?....or how should you approach this limit?

Thank you.
$\displaystyle \forall{x}\in\mathbb{R}\quad{-1\leq\cos(x)\leq{1}}$

$\displaystyle \therefore\quad\forall{x}\in\mathbb{R}\quad{\frac{-1}{e^x}\leq\frac{\cos(x)}{e^x}\leq\frac{1}{e^x}}$

$\displaystyle \therefore\quad\lim_{x\to\infty}\frac{1}{e^x}\leq\ lim_{x\to\infty}\frac{\cos(x)}{e^x}\leq\lim_{x\to\ infty}\frac{1}{e^x}$

$\displaystyle \therefore\quad{0}\leq\lim_{x\to\infty}\frac{\cos( x)}{e^x}\leq{0}$

$\displaystyle \Rightarrow\lim_{x\to\infty}\frac{\cos(x)}{e^x}=0$

Also you could consider that cosine is just an infnite series of polynomials which will be asymptotically equaivalent to its largest term which no matter how big is of a lower order infinity than the exponential function, thus the limit is zero.

3. Originally Posted by Mathstud28
$\displaystyle \forall{x}\in\mathbb{R}\quad{-1\leq\cos(x)\leq{1}}$

$\displaystyle \therefore\quad\forall{x}\in\mathbb{R}\quad{\frac{-1}{e^x}\leq\frac{\cos(x)}{{\color{red}e}^x}\leq\fr ac{1}{e^x}}$

$\displaystyle \therefore\quad\lim_{x\to\infty}\frac{1}{e^x}\leq\ lim_{x\to\infty}\frac{\cos(x)}{e^x}\leq\lim_{x\to\ infty}\frac{1}{e^x}$

$\displaystyle \therefore\quad{0}\leq\lim_{x\to\infty}\frac{\cos( x)}{e^x}\leq{0}$

$\displaystyle \Rightarrow\lim_{x\to\infty}\frac{\cos(x)}{e^x}=0$
I think I fixed a tiny error in red...

--Chris

4. Originally Posted by Chris L T521
I think I fixed a tiny error in red...

--Chris
Actually, I think you made it incorrect.

5. Originally Posted by RedBarchetta
$\displaystyle \mathop {\lim }\limits_{x \to \infty } \frac{{\cos x}} {{e^x }} = 0$

L'Hopitals rule really doesn't help in this situation. How would you determine this? As x approaches infinity, the limit of cosine is indefinite. Also as x approaches infinity, e^x approaches infinity.

So basically, are you supposed to look at it from the perspective that no matter what is going on with cosine, e^x becomes so massive that the numerator is negligible?....or how should you approach this limit?

Thank you.
The limit of a quotient is equal to the quotient of the limits.

We know that $\displaystyle \cos{x}$ oscillates between -1 and 1. So as $\displaystyle x \to \infty$ we still have $\displaystyle \cos{x}$ oscillating between these values.

As $\displaystyle x \to \infty, e^x \to \infty$. This means that the denominator gets HUGE. Divide any number by something very big, you get something very small.

So I would think that the limit tends to 0, as the denominator getting huge makes the quotient very small.

6. Originally Posted by Mathstud28
Actually, I think you made it incorrect.
No, he fixed it. You divided the whole inequality by $\displaystyle e^x$, not x.

7. Originally Posted by Prove It
No, he fixed it. You divided the whole inequality by $\displaystyle e^x$, not x.
I know this...that was my intention...

8. ......So could you put this into layman's terms? Is this just the sandwich theorem?

9. Originally Posted by RedBarchetta
......So could you put this into layman's terms? Is this just the sandwich theorem?
Exactly...that is all it is...the fancy notation is just to make me look smart

10. Originally Posted by RedBarchetta
......So could you put this into layman's terms? Is this just the sandwich theorem?
Yes, Mathstud used the sandwich theorem.

I just used the fact that the denominator would get huge implies that the quotient gets small.

Either way, you get the limit = 0.

11. Originally Posted by Mathstud28
Exactly...that is all it is...the fancy notation is just to make me look smart
Geez...stop trying to make yourself look smart..