# Does the Limit exist?

1. May 25, 2013

### mathgeek69

1. lim as x->∞of 3x3+cosx/sinx-x3

2. My Attempt below:

3. lim as x->∞ of 3x3+lim as x->∞ of cosx/lim as x->∞ of sinx-lim as x->∞ of x3

Breaking down problem even further:

lim as x->∞ of 3x3 = ∞
lim as x->∞ of cosx= 1. This is because by squeeze theorem, -1<cosx<1, therefore lim reaches 1
lim as x->∞ of sinx= 1. This is because by squeeze theorem, -1<sinx<1, therefore lim reaches 1
lim as x->∞ of x3 =∞

∴ = +1/1-∞ = -∞

Is this correct?

2. May 25, 2013

### Mandelbroth

I don't think you understand the squeeze theorem. Do you mean $\frac{3x^3+\cos{x}}{\sin{x}-x^3}$ or $3x^3 + \frac{\cos{x}}{\sin{x}-x^3}$?

3. May 25, 2013

### mathgeek69

$\frac{3x^3+\cos{x}}{\sin{x}-x^3}$

4. May 25, 2013

### janhaa

$\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}$
the limit does exist

5. May 25, 2013

### mathgeek69

And does the limit = -∞

6. May 25, 2013

### janhaa

no, sorry, it's

$\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}=-3$

try with the squeeze theorem

7. May 25, 2013

### mathgeek69

I am not getting the concept of squeeze theorem clearly. I thought the following:

lim as x->∞ of cosx= 1. This is because by squeeze theorem, -1<cosx<1, therefore lim reaches 1
lim as x->∞ of sinx= 1. This is because by squeeze theorem, -1<sinx<1, therefore lim reaches 1

What am I doing wrong?

8. May 25, 2013

### Mandelbroth

You aren't understanding what the squeeze rule says.

Let $g(x)\leq f(x)\leq h(x)$. The squeeze rule says that, if $\displaystyle \lim_{x\rightarrow\alpha}g(x)=\lim_{x\rightarrow\alpha}h(x)=L$, then $\displaystyle \lim_{x\rightarrow\alpha}f(x)=L$.

For example, the squeeze theorem doesn't apply for $\displaystyle \lim_{x\rightarrow\infty}\sin{x}$ because $\displaystyle \lim_{x\rightarrow\infty}[-1]\neq\lim_{x\rightarrow\infty}[1]$.

Edit:
Think of it this way.

$$\lim_{x\rightarrow\infty}\frac{3x^3+\cos{x}}{\sin{x}-x^3}=\lim_{x\rightarrow\infty}\frac{3x^3(1+\frac{\cos{x}}{3x^3})}{-x^3(-\frac{\sin{x}}{x^3}+1)}$$

Last edited: May 25, 2013
9. May 25, 2013

### mathgeek69

ok then. When applying the squeeze theorem,

how are you concluding to this?

$\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}=-3$

10. May 25, 2013

### mathgeek69

so ...

$$\lim_{x\rightarrow\infty}\frac{3x^3+\cos{x}}{\sin{x}-x^3}=\lim_{x\rightarrow\infty}\frac{3x^3(1+\frac{\cos{x}}{3x^3})}{-x^3(-\frac{\sin{x}}{x^3}+1)}$$ = -3

Cheers,

11. May 26, 2013

### HallsofIvy

Staff Emeritus
I'm glad you got it. Now, let me point out that $\lim_{x\to \infty} sin(x)$ and $\lim_{x\to \infty} cos(x)$ are NOT 0. They continue to alternate between -1 and 1 and do not "approach" any number. You use the "squeeze theorem" to argue that because $-1\le sin(x)\le 1$ and $-1\le cos(x)\le 1$, $-\frac{1}{x}\le \frac{sin(x)}{x}\le \frac{1}{x}$ and $-\frac{1}{x}\le \frac{sin(x)}{x}\le \frac{1}{x}$ so that $\lim_{x\to\infty} \frac{sin(x)}{x}= \lim_{x\to\infty} \frac{cos(x)}{x}= 0$.