What is the limit of (3x³ + cos(x)) / (sin(x) - x³) as x approaches infinity?

[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?

 

[Mandelbroth]

1. lim as x->∞of 3x³+cosx/sinx-x³2. 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?

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

 

[mathgeek69]

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

 

[janhaa]

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

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

 

[mathgeek69]

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

And does the limit = -∞

 

[janhaa]

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

no, sorry, it's

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

try with the squeeze theorem

 

[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?

 

[Mandelbroth]

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?

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.

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

 

[mathgeek69]

ok then. When applying the squeeze theorem,

how are you concluding to this?$\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}=-3$

 

[mathgeek69]

Edit:
Think of it this way.

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

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

I GET IT! @Mandelbroth. your new edit was helpful! VERY HELPFUL

Cheers,

 

[HallsofIvy]

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 $\lim_{x\to\infty} \frac{sin(x)}{x}= \lim_{x\to\infty} \frac{cos(x)}{x}= 0$.