# Homework Help: Limits homework

1. Sep 12, 2009

### step1536

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

IF 5X<=f(X)<=X^3+2
FIND LIM F(X)IS X=2

2. Sep 12, 2009

Re: Limits

I'm sorry, what are you asking? Is it this:

If $5x\le f(x)\le x^3+2$ then find $\lim_{x\rightarrow2}f(x)$ ?

3. Sep 12, 2009

### step1536

Re: Limits

yes that is correct

4. Sep 12, 2009

Re: Limits

What do you know about the limit of a function that is being "Sandwiched" or "Squeezed" between two other functions? hint...hint...

5. Sep 12, 2009

### step1536

Re: Limits

so I have 10<=2<=10 and the equations equals 10

6. Sep 12, 2009

Re: Limits

Yes.

7. Sep 12, 2009

### step1536

Re: Limits

I think I understand limits but this last question says that lim x=0
evaluate limx^6 cos 7/x
I came up with 0 or would it be limit does not exist.

8. Sep 12, 2009

Re: Limits

$$\lim_{x\rightarrow0}x^6\cos(\frac{7}{x})$$ ?

9. Sep 12, 2009

### step1536

Re: Limits

yes that is correct

10. Sep 12, 2009

Re: Limits

Hmmm... I am not sure. Perhaps someone else can chime in here. I don't remember all of the 'tricks' to evaluating limits.

If you can find a way to rewrite the function such that the 'x' in the denominator can be cancelled, then the limit exists.

I would play around with the trig identities maybe?

11. Sep 13, 2009

### snipez90

Re: Limits

The limit should be 0, since x gets arbitrarily small. Although there is oscillation from the cos, the function is bounded. You should try proving this rigorously.

12. Sep 13, 2009

### VietDao29

Re: Limits

No.. you don't need any trig identity here. When dealing with limits of this kind, i.e involving the product of one part tending to 0, and the other part oscillating. Then, one should think right about the range of the oscillating part, and how "big" can it get.

Hint hint..

13. Sep 13, 2009

Re: Limits

Is one of the 'factors' oscillating though? It looks undefined. But, like I said, it's been awhile

$$\lim_{x\rightarrow0}x^6\cos(\frac{7}{x})$$

14. Sep 13, 2009

### VietDao29

Re: Limits

The function is undefined at x = 0. However the limit does exist when x tends to 0.

Another example is:

$$\frac{x ^ 2}{x}$$ is undefined at x = 0. But:

$$\lim_{x \rightarrow 0} \frac{x ^ 2}{x} = \lim_{x \rightarrow 0} x = 0$$.

------------------

Or:

$$\frac{\sin(x)}{x}$$ is undefined at x = 0. But:

$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$.

Because, when taking limit as x tends to 0. We just consider the values for x near 0 (0 excluded). So, when the function is undefined at x = 0, the limit may still exist. :)

Well, if you are still unsure about how to do it. I'll give you a hint then:

1. Find the range for cos(x)
2. Then, try to apply the "Sandwich Theorem" like this:
$$... \leq x ^ 6 \cos\left(\frac{7}{x} \right) \leq ...$$

15. Sep 13, 2009