Evaluating limits

[ScienceJen]

I'm working through many questions and I just want to see if I'm on the right track. For a., x is approaching 0 from the left and is therefore less than 0.
So, f(x) = 2/x in this case. Do I need to elaborate? I think this function is discontinuous, since the f(x) 1/x is two lines (one in quadrant 1 and one in quadrant 3).

 

[BigBeachBanana]

View attachment 31879
I'm working through many questions and I just want to see if I'm on the right track. For a., x is approaching 0 from the left and is therefore less than 0.
So, f(x) = 2/x in this case. Do I need to elaborate? I think this function is discontinuous, since the f(x) 1/x is two lines (one in quadrant 1 and one in quadrant 3).

It would beneficial for you to look at the graph.

 

[JeffM]

Your answer is correct, but I am not sure your explanation is convincing.

2/x is discontinuous at x = 0 for the exact same reason that 1/x is discontinuous at x = 0. But you have not given that reason.

The reason is that 2/x grows without bound as x approaches 0. How would you show that mathematically?

 

[ScienceJen]

Your answer is correct, but I am not sure your explanation is convincing.

2/x is discontinuous at x = 0 for the exact same reason that 1/x is discontinuous at x = 0. But you have not given that reason.

The reason is that 2/x grows without bound as x approaches 0. How would you show that mathematically?

Thank you!
f(2/x) as x approaches 0 = (- infinity, infinity). ? (using the infinity symbol, not writing the word out).

 

[ScienceJen]

View attachment 31879
I'm working through many questions and I just want to see if I'm on the right track. For a., x is approaching 0 from the left and is therefore less than 0.
So, f(x) = 2/x in this case. Do I need to elaborate? I think this function is discontinuous, since the f(x) 1/x is two lines (one in quadrant 1 and one in quadrant 3).

For b., as x approaches zero from the right, I would say f(x) = sqrt x + cos(pi/x). Therefore, wouldn't the x's in the formula become 0 and leave us with cos? That doesn't make sense.

 

[JeffM]

Thank you!
f(2/x) as x approaches 0 = (- infinity, infinity). ? (using the infinity symbol, not writing the word out).

Let’s talk about mathematical notation. $(- \infty, \ \infty)$ means every real number. That is not what you mean. You mean $\lim_{x \rightarrow 0} \dfrac{1}{x} = \infty \text { or } - \infty.$

Do you understand the distinction? A limit is NEVER an interval.

But infinity is not a real number. So what in the world does a statement like $\lim_{x \rightarrow 0}\dfrac{1}{x^2} = \infty$ mean. It certainly is not like the statement $\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1$, which identifies a real number. In a very real sense, the statement that a limit is infinite is a contradiction. We are saying no real limit exists.

$$\lim_{x \rightarrow a} f(x) = \infty \iff \\ \text {For ANY positive real number } M, \ \exists \ \delta > 0 \text { such that }\\ M < f(x) \text { if } 0 < |a - x| < \delta.$$
In other words, if x gets close enough to a, then f(x) becomes bigger than any positive real number imaginable.

 

[ScienceJen]

Let’s talk about mathematical notation. $(- \infty, \ \infty)$ means every real number. That is not what you mean. You mean $\lim_{x \rightarrow 0} \dfrac{1}{x} = \infty \text { or } - \infty.$

Do you understand the distinction? A limit is NEVER an interval.

But infinity is not a real number. So what in the world does a statement like $\lim_{x \rightarrow 0}\dfrac{1}{x^2} = \infty$ mean. It certainly is not like the statement $\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x} = 1$, which identifies a real number. In a very real sense, the statement that a limit is infinite is a contradiction. We are saying no real limit exists.

$$\lim_{x \rightarrow a} f(x) = \infty \iff \\ \text {For ANY positive real number } M, \ \exists \ \delta > 0 \text { such that }\\ M < f(x) \text { if } 0 < |a - x| < \delta.$$
In other words, if x gets close enough to a, then f(x) becomes bigger than any positive real number imaginable.

Thank you for explaining that. I'm just learning about delta and epsilon. If la-xl is less than delta (and delta is a very small positive number close to zero), then how can f(x) be an unimaginably large number?

 

[JeffM]

For b., as x approaches zero from the right, I would say f(x) = sqrt x + cos(pi/x). Therefore, wouldn't the x's in the formula become 0 and leave us with cos? That doesn't make sense.

First off, you have misread the definition of the function.

$$0 < x \le 5 \implies f(x) = \sqrt{x} + \cos(\pi x) \ne \sqrt{5} + \cos \left ( \dfrac{\pi}{x} \right ).$$

 

[Steven G]

View attachment 31879
I'm working through many questions and I just want to see if I'm on the right track. For a., x is approaching 0 from the left and is therefore less than 0.
So, f(x) = 2/x in this case. Do I need to elaborate? I think this function is discontinuous, since the f(x) 1/x is two lines (one in quadrant 1 and one in quadrant 3).

Yes you do need to elaborate more. Even if everything you said is true, you never said what the limit equals!!!! No one asked you if the function was continuous or not!

 

[Steven G]

Thank you for explaining that. I'm just learning about delta and epsilon. If la-xl is less than delta (and delta is a very small positive number close to zero), then how can f(x) be an unimaginably large number?

delta x does not have to be a very small positive number. The size of delta depends on the size of epsilon.

 

[ScienceJen]

Yes you do need to elaborate more. Even if everything you said is true, you never said what the limit equals!!!! No one asked you if the function was continuous or not!

Ok. Yeah, taking calculus without an instructor is very difficult for me. The question is asking to evaluate the limits and if a limit does not exist, explain why. I think your Darwin quote is apt, by the way. That's how I feel right now.
So, in evaluating the limit, I'm suppose to state infinity or not...or solve for x in a way? At the end of the limit examples, it asks to state where is the function not continuous - I don't know that yet. Obviously, part a is continuous if it increases without bounds?

First off, you have misread the definition of the function.

$$0 < x \le 5 \implies f(x) = \sqrt{x} + \cos(\pi x) \ne \sqrt{5} + \cos \left ( \dfrac{\pi}{x} \right ).$$

I was thinking it would actually be 0 in place of the x's, since x is approaching 0 from the right. ?

As for part c), limit of f(x) as x approaches 5 from the left = sqrt x/cos(pi*x) = Do I plug 5 in to replace the x's?

Part d) = x/(x-5) = 5/(5-5) = 5/0 = 0. Limit DNE when denominator = 0. ?

Thank you again. I know you are probably finding this as frustrating as I am.

 

[JeffM]

Thank you for explaining that. I'm just learning about delta and epsilon. If la-xl is less than delta (and delta is a very small positive number close to zero), then how can f(x) be an unimaginably large number?

That is a great question, but it is late, I am sleepy and have had a couple of glasses of wine. I’ll answer tomorrow if I can sign in. (I could not sign in for six days before today and for at least an hour today. No idea what is preventing my logging on.)

$$\lim_{x \rightarrow 0} \cos ( \pi x) = 0.$$

 

[JeffM]

That is a great question, but it is late, I am sleepy and have had a couple of glasses of wine. I’ll answer tomorrow if I can sign in. (I could not sign in for six days before today and for at least an hour today. No idea what is preventing my logging on.)

$$\lim_{x \rightarrow 0} \cos ( \pi x) = 0.$$

That should say

$$\lim_{x \rightarrow 0} \cos ( \pi x) = 1.$$

 

[JeffM]

If la-xl is less than delta (and delta is a very small positive number close to zero), then how can f(x) be an unimaginably large number?

That is a really intelligent question.

I am not going to give a formal answer. Instead, I am going to give two informal answers that may be more intuitive.

Consider the function $f(x) = \dfrac{1}{x^4}$. We want to see what happens as x approaches more closely to zero.

$$0 < |x| < 10^{-1} = \dfrac{1}{10} \implies 0 < x^4 < \dfrac{1}{10000} \implies \dfrac{1}{x^4} > 10000\\ 0 < |x| < 10^{-2} = \dfrac{1}{100} \implies 0 < x^4 < \dfrac{1}{100,000,000} \implies \dfrac{1}{x^4} > 100,000,000.$$
In the second line, x is closer to zero than in the first line, but the minimum possible value of the function is much greater.

More generally, but still not rigorously,

$$0 < \dfrac{1}{\alpha} < \dfrac{1}{\beta} \implies \alpha > 0 \text { and } \beta > 0 \implies \alpha \beta > 0.\\ \therefore 0 < \dfrac{1}{\alpha} < \dfrac{1}{\beta} \implies \alpha \beta * 0 < \alpha \beta * \dfrac{1}{\alpha} < \alpha \beta * \dfrac{1}{\beta} \implies\\ 0 < \beta < \alpha.$$
Similarly

$$\dfrac{1}{\alpha} < \dfrac{1}{\beta} < 0\implies \alpha < 0 \text { and } \beta < 0 \implies \alpha \beta > 0.\\ \therefore \dfrac{1}{\alpha} < \dfrac{1}{\beta} < 0 \implies \alpha \beta * \dfrac{1}{\alpha} < \alpha \beta * \dfrac{1}{\beta} <\alpha \beta * 0 \implies\\ \beta < \alpha < 0.$$
The reciprocal of what is closer to zero is farther from zero.

A lot of this limit stuff is a very precise way to state what is intuitively simple.

 

[ScienceJen]

That should say

$$\lim_{x \rightarrow 0} \cos ( \pi x) = 1.$$

Thank you Jeff. I have a few things written down from the e-text of my course, but it doesn't include everything and it's very frustrating.

 

[ScienceJen]

That is a really intelligent question.

I am not going to give a formal answer. Instead, I am going to give two informal answers that may be more intuitive.

Consider the function $f(x) = \dfrac{1}{x^4}$. We want to see what happens as x approaches more closely to zero.

$$0 < |x| < 10^{-1} = \dfrac{1}{10} \implies 0 < x^4 < \dfrac{1}{10000} \implies \dfrac{1}{x^4} > 10000\\ 0 < |x| < 10^{-2} = \dfrac{1}{100} \implies 0 < x^4 < \dfrac{1}{100,000,000} \implies \dfrac{1}{x^4} > 100,000,000.$$
In the second line, x is closer to zero than in the first line, but the minimum possible value of the function is much greater.

More generally, but still not rigorously,

$$0 < \dfrac{1}{\alpha} < \dfrac{1}{\beta} \implies \alpha > 0 \text { and } \beta > 0 \implies \alpha \beta > 0.\\ \therefore 0 < \dfrac{1}{\alpha} < \dfrac{1}{\beta} \implies \alpha \beta * 0 < \alpha \beta * \dfrac{1}{\alpha} < \alpha \beta * \dfrac{1}{\beta} \implies\\ 0 < \beta < \alpha.$$
Similarly

$$\dfrac{1}{\alpha} < \dfrac{1}{\beta} < 0\implies \alpha < 0 \text { and } \beta < 0 \implies \alpha \beta > 0.\\ \therefore \dfrac{1}{\alpha} < \dfrac{1}{\beta} < 0 \implies \alpha \beta * \dfrac{1}{\alpha} < \alpha \beta * \dfrac{1}{\beta} <\alpha \beta * 0 \implies\\ \beta < \alpha < 0.$$
The reciprocal of what is closer to zero is farther from zero.

A lot of this limit stuff is a very precise way to state what is intuitively simple.

Thank you. Learning calculus for someone who is extremely analytical is very difficult. I like to know the how's and why's. At first the reciprocal explanation didn't make sense, but after re and re-reading it I think I understand for the most part. I wish that something so simple didn't have to be so complicated.

 

[JeffM]

Thank you. Learning calculus for someone who is extremely analytical is very difficult. I like to know the how's and why's. At first the reciprocal explanation didn't make sense, but after re and re-reading it I think I understand for the most part. I wish that something so simple didn't have to be so complicated.

The problem with the way calculus is taught is that generating proofs is extremely difficult. It took mathematicians 200 years to do so. Did you know that Newton used calculus to generate results but used geometry in his proofs? In the eighteenth century, calculus was sort of a blend of physical experimentation and intuitive reasoning. It was not until the early 19th century that Cauchy started to put calculus on a logically secure basis. We try to teach 200 years of concentrated thought in just a week or so. No wonder students are perplexed. Applying calculus is MUCH easier than developing a logically rigorous framework for it.

 

[Deleted member 4993]

5/(5-5) = 5/0 = 0

That is incorrect

5/(5-5) = 5/0 = _∞ (or DNE)

 

[Steven G]

Thank you!
f(2/x) as x approaches 0 = (- infinity, infinity). ? (using the infinity symbol, not writing the word out).

Actually f(2/x) approaches 0 as x approaches 0^- and f(2/x) approaches 1 as x approaches 0⁺. Do you see this??

 

[Deleted member 4993]

Actually f(2/x) approaches 0 as x approaches 0^- and f(2/x) approaches 1 as x approaches 0⁺. Do you see this??

Steven,

Are you still discussing f(x) in OP?