Confused with the terms used??

[MathNoob94]

Hi, so pretty much I've read this example about 50 times and still do not get what it means and how you get from one term to the next.

So in the picture provided there is an example f(x) = 1/1-e^1/x and the teacher has wrote on notes x--> 0 (left side) and 1/x --> -infinity (I get up to here). then e^1/x --> 0(right side). If anyone can, can you please explain the steps to evaluating a limit and explain what my professor means. On a test will I be required to show all this? I can't understand why I find this so difficult.


Thank You in advance!

 

[stapel]

Hi, so pretty much I've read this example about 50 times and still do not get what it means and how you get from one term to the next.

I'm seeing a full page of examples, with handwriting all over it. Which is "this example"? Which of the "wrote on notes" are causing the confusion?

Please reply with the text of the one exercise, along with the handwritten notes which apply (in sequential order!). Thank you.

 

[ksdhart]

What I'm getting from this is that you're confused by this step right here:

$\displaystyle \displaystyle \lim _{x\to 0^-}\left(e^{\frac{1}{x}}\right)=0$

Think about what the limit actually means. It's the value that the function approaches as x gets closer and closer to the target value (in this case 0). So think about what happens as to the function e^(1/x) as x gets incredibly small. Note that because x is approaching 0 from below, it is always negative. Note what happens to (1/x) as x approaches 0. Then think about what would happen if you raised e to that power. Making a table of these values might help you a lot, too.

 

[Steven G]

Hi, so pretty much I've read this example about 50 times and still do not get what it means and how you get from one term to the next.

So in the picture provided there is an example f(x) = 1/1-e^1/x and the teacher has wrote on notes x--> 0 (left side) and 1/x --> -infinity (I get up to here). then e^1/x --> 0(right side). If anyone can, can you please explain the steps to evaluating a limit and explain what my professor means. On a test will I be required to show all this? I can't understand why I find this so difficult.


Thank You in advance!
View attachment 5339

You need to follow the composition of functions. You are told that as x goes to 0+ that (1/x) goes to infinity. Think about the sign first. The numerator is 1 and is clearly positive. The denominator is getting closer and closer to ) but always positive. So the fraction being +/+ will be positive. Compute the value of the entries in the following sequence: 1/1, 1/(1/2), 1/(1/3), 1/(1/100), 1/(1/100000). Note the the denominators are getting smaller and smaller, ie getting closer and closer to 0. So as x->0+, 1/x approaches infinity. So e^(1/x) approaches e^infinity or infinity. Now as x->0+, 1+e^(1/x) approaches 1+infinity=infinity. Now as x->0+, 1/(1+e^(1/x))=1/infinity=0. Since the numerator is 1, ie positive, and the denominator is approaching infinity, ie positive we have the quotient positive. So actually the lim is 0+/.

Use the same logic for x->0-

 

[MathNoob94]

What I'm getting from this is that you're confused by this step right here:

$\displaystyle \displaystyle \lim _{x\to 0^-}\left(e^{\frac{1}{x}}\right)=0$

Think about what the limit actually means. It's the value that the function approaches as x gets closer and closer to the target value (in this case 0). So think about what happens as to the function e^(1/x) as x gets incredibly small. Note that because x is approaching 0 from below, it is always negative. Note what happens to (1/x) as x approaches 0. Then think about what would happen if you raised e to that power. Making a table of these values might help you a lot, too.

Sorry! It's example 6 the top half. Yes I am confused with that. Even when the teacher goes step by step explaining it I don't understand how the steps to approach that limit ex where it says x approaches 0 and etc

 

[MathNoob94]

You need to follow the composition of functions. You are told that as x goes to 0+ that (1/x) goes to infinity. Think about the sign first. The numerator is 1 and is clearly positive. The denominator is getting closer and closer to ) but always positive. So the fraction being +/+ will be positive. Compute the value of the entries in the following sequence: 1/1, 1/(1/2), 1/(1/3), 1/(1/100), 1/(1/100000). Note the the denominators are getting smaller and smaller, ie getting closer and closer to 0. So as x->0+, 1/x approaches infinity. So e^(1/x) approaches e^infinity or infinity. Now as x->0+, 1+e^(1/x) approaches 1+infinity=infinity. Now as x->0+, 1/(1+e^(1/x))=1/infinity=0. Since the numerator is 1, ie positive, and the denominator is approaching infinity, ie positive we have the quotient positive. So actually the lim is 0+/.

Use the same logic for x->0-

Thank You so much! looking at the way you explained it makes it easier to understand however, just a quick question why "Now as x->0+, 1/(1+e^(1/x))=1/infinity=0" is the equation equal to zero?

 

[MathNoob94]

You need to follow the composition of functions. You are told that as x goes to 0+ that (1/x) goes to infinity. Think about the sign first. The numerator is 1 and is clearly positive. The denominator is getting closer and closer to ) but always positive. So the fraction being +/+ will be positive. Compute the value of the entries in the following sequence: 1/1, 1/(1/2), 1/(1/3), 1/(1/100), 1/(1/100000). Note the the denominators are getting smaller and smaller, ie getting closer and closer to 0. So as x->0+, 1/x approaches infinity. So e^(1/x) approaches e^infinity or infinity. Now as x->0+, 1+e^(1/x) approaches 1+infinity=infinity. Now as x->0+, 1/(1+e^(1/x))=1/infinity=0. Since the numerator is 1, ie positive, and the denominator is approaching infinity, ie positive we have the quotient positive. So actually the lim is 0+/.

Use the same logic for x->0-

Okay. would this be correct for x -> 0-

As x->0-, 1/x -> -infinity, as x-> 0-, e^1/x --> 0+, as x -> 0-, 1+e^1/x -> 1+, as x -> 0- 1/1+e^1/x -> 1- (negative because x -> 0-) (I entered in 1/1+e^1/.5 so on and saw that overall expression was getting closer to 1 from positive) therefore, overall expression will be.
lim 1/1+e^1/x = 1
x->0

******
The thing I believe I was getting confused was what values i should enter in for x to see how the equation responds. I just realized I should enter in values to what x approaches! It makes SO! MUCH MORE SENSE!!!! Thank You and Thank You to Everyone else who explained this example!!!
*******