Limits and Vertical Asymptotes

[knowLittle]

Homework Statement

I found that the domain of f was defined for all reals x<-1/3 and x>1/3. Now, I need to find the limits and vertical asymptotes of
$\lim _{\chi \rightarrow \dfrac {1} {3}}\dfrac {x+2} {\sqrt {9x^{2}-1}}$

According to Wolfram Alpha, there is no limit. But, I found that the limit approaches 54(80/9)^(1/2).

The Attempt at a Solution

$\dfrac {\dfrac {x} {x}+\dfrac {2} {x}} {\dfrac {\sqrt {9x^{2}-1}} {\sqrt {x^{2}}}}=\dfrac {1+\dfrac {2} {x}} {\sqrt {9-\dfrac {1} {x^{2}}}}$
Then, I use L' Hospital:

$\dfrac {-2x^{-2}} {\dfrac {1} {2}\left( 9-x^{2}\right) ^{-1 / 2}\left( -2x\right) }=\dfrac {2} {x^{3}}\sqrt {9-x^{2}}$

So, the limit approaches 54 *(80/9)^1/2

Please, help.

 

[SammyS]

Homework Statement

I found that the domain of f was defined for all reals x<-1/3 and x>1/3. Now, I need to find the limits and vertical asymptotes of
$\lim _{\chi \rightarrow \dfrac {1} {3}}\dfrac {x+2} {\sqrt {9x^{2}-1}}$

According to Wolfram Alpha, there is no limit. But, I found that the limit approaches 54(80/9)^(1/2).

The Attempt at a Solution

$\dfrac {\dfrac {x} {x}+\dfrac {2} {x}} {\dfrac {\sqrt {9x^{2}-1}} {\sqrt {x^{2}}}}=\dfrac {1+\dfrac {2} {x}} {\sqrt {9-\dfrac {1} {x^{2}}}}$
Then, I use L' Hospital:

L'Hôpital's rule can't be applied here.

This limit is not of the form 0/0 or ∞/∞ .

$\dfrac {-2x^{-2}} {\dfrac {1} {2}\left( 9-x^{2}\right) ^{-1 / 2}\left( -2x\right) }=\dfrac {2} {x^{3}}\sqrt {9-x^{2}}$

So, the limit approaches 54 *(80/9)^1/2

Please, help.

 

[knowLittle]

You are right, but how can I proceed?

 

[STEMucator]

Try multiplying by 1.

 

[knowLittle]

I am trying the squeeze theorem now, but still no answer. There have to be some sort of crafty manipulation to the equation. It's very easy to find the limit at infinity, though.

 

[STEMucator]

Hmm now that I took a closer look at this. You should get no limit at all. Think about what happens as you approach 1/3 from the right and 1/3 from the left.

 

[knowLittle]

I know that the function grows without bound as x approaches x=-1/3 and x=1/3, but I need to prove it.

 

[Mark44]

I am trying the squeeze theorem now, but still no answer. There have to be some sort of crafty manipulation to the equation. It's very easy to find the limit at infinity, though.

Easy, but not relevant.

Hmm now that I took a closer look at this. You should get no limit at all. Think about what happens as you approach 1/3 from the right and 1/3 from the left.

Since the domain is (-∞, -1/3) U (1/3, ∞) it makes no sense to approach 1/3 from the left. The function is not defined on [-1/3, 1/3].

The problem should be restated as this:
$$ \lim_{x \to 1/3^+}\frac{x + 2}{\sqrt{9x^2 - 1}}$$

The limit can be evaluated directly by noting what the numerator is doing and what the denominator is doing as x gets close to 1/3 from the right.

 

[knowLittle]

I see that the numerator is approaching 2.333... and the denominator approaches 0.
I know that, when lim x-> 0 1/(x^2) it's very clear that the function grows without bounds, but in this occasion I just can't see how it grows.

 

[SammyS]

I see that the numerator is approaching 2.333... and the denominator approaches 0.
I know that, when lim x-> 0 1/(x^2) it's very clear that the function grows without bounds, but in this occasion I just can't see how it grows.

Yes, the denominator approaches 0, but from what direction?

Is the denominator ever negative?

 

[Mark44]

I see that the numerator is approaching 2.333... and the denominator approaches 0.

Yes, the denominator approaches 0, but from what direction?

The answer to Sammy's question is crucial to determining what the limit is.

 

[knowLittle]

The denominator is never negative. It approaches 0 from the right hand side.

Is this the same as saying: (any constant)/ (a number infinitely small that never really reaches zero) ?
Then, it follows by the same principle as 1/ (x^2) as x->0 grows without bound?

 

[Mark44]

The denominator is never negative. It approaches 0 from the right hand side.

Is this the same as saying: (any constant)/ (a number infinitely small that never really reaches zero) ?

I think you have the idea, but you're not saying it very well.

If the numerator is a positive constant, and the denominator approaches zero but remains constant, then the right-hand limit of this quotient is infinity.

Then, it follows by the same principle as 1/ (x^2) as x->0 grows without bound?

 

[knowLittle]

What do you mean by, "the denominator approaches zero, but remains constant"?
How do you know that it remains constant? How so?

 

[SammyS]

What do you mean by, "the denominator approaches zero, but remains constant"?
How do you know that it remains constant? How so?

I think Mark meant to say

the denominator approaches zero, but remains positive.​

 

[knowLittle]

Thank you very much to all. Now, it remains clear that the same principle of limit is there.

Thank you, again!