Finding the limit of f(x) as x approaches 2 given 5x ≤ f(x) ≤ x³+2

[step1536]

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

 

[Saladsamurai]

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

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

 

[step1536]

yes that is correct

 

[Saladsamurai]

yes that is correct

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

 

[step1536]

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

 

[Saladsamurai]

Yes.

 

[step1536]

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.

 

[Saladsamurai]

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

 

[step1536]

yes that is correct

 

[Saladsamurai]

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?

 

[snipez90]

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.

 

[VietDao29]

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?

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..

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

 

[Saladsamurai]

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..

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

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

 

[VietDao29]

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

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

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:
   $$... \leq x ^ 6 \cos\left(\frac{7}{x} \right) \leq ...$$

 

[Saladsamurai]

Ah. I see, I am actually just reviewing limits fir the first time in years. I have past the section on how to deal with rational functions with zero denominators if the zero in the denominator can be 'eliminated.'

But, have yet to get to the part where they cannot.

Thanks for the preview