TT's Maths Thread

[TrueTears]

1. Is there a way to evaluate without using the squeeze theorem?

2. Also how to evaluate ? (How to set it out formally?)

Do you just say since approaches , then where will approach ?

3. Also how do I show and are the vertical asymptotes of ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have

Clearly is a vertical asymptote.

But why?

We can show that

This is because if we take a value right next to the right of then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to , the limit is .



This is because if we take a value right next to the left of then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to , the limit is .

Now how do we apply the same "technique" to the question above?

Thanks

[humph]

1. Is there a way to evaluate without using the squeeze theorem?

Probably, though I don't see why you would, seeing as the squeeze theorem is the natural way to evaluate these kind of limits. When you want to show something tends to zero, it's often easiest to just compare it to some other functions that also go to zero, and that the comparison forces the original function to have the same limit - that's basically the squeeze theorem (or sandwich theorem, or sandwich rule, or whatever you call it).

2. Also how to evaluate ? (How to set it out formally?)

Do you just say since approaches , then where will approach ?

I suppose, though that might not be rigorous enough for some people (in which case you could do an argument, but surely that'd be overkill). One other trick to remember is to change variables/write a function as a composition of functions.

3. Also how do I show and are the vertical asymptotes of ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have

Clearly is a vertical asymptote.

But why?

We can show that

This is because if we take a value right next to the right of then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to , the limit is .



This is because if we take a value right next to the left of then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to , the limit is .

Now how do we apply the same "technique" to the question above?

Thanks

Well is a vertical asymptote because the denominator tends to zero as tends to , while the numerator remains bounded away from zero as tends to . You can again make some intelligent change of variables to show that it comes down to the fact that has a vertical asymptote at if for all in some interval containing .

[moekamo]

3. Also how do I show and are the vertical asymptotes of ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have

Clearly is a vertical asymptote.

But why?

We can show that

This is because if we take a value right next to the right of then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to , the limit is .



This is because if we take a value right next to the left of then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to , the limit is .

Now how do we apply the same "technique" to the question above?

Thanks

cant you just divide so , then use the same method?

[TrueTears]

1. Is there a way to evaluate without using the squeeze theorem?

Probably, though I don't see why you would, seeing as the squeeze theorem is the natural way to evaluate these kind of limits. When you want to show something tends to zero, it's often easiest to just compare it to some other functions that also go to zero, and that the comparison forces the original function to have the same limit - that's basically the squeeze theorem (or sandwich theorem, or sandwich rule, or whatever you call it).

2. Also how to evaluate ? (How to set it out formally?)

Do you just say since approaches , then where will approach ?

I suppose, though that might not be rigorous enough for some people (in which case you could do an argument, but surely that'd be overkill). One other trick to remember is to change variables/write a function as a composition of functions.

3. Also how do I show and are the vertical asymptotes of ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have

Clearly is a vertical asymptote.

But why?

We can show that

This is because if we take a value right next to the right of then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to , the limit is .



This is because if we take a value right next to the left of then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to , the limit is .

Now how do we apply the same "technique" to the question above?

Thanks

Well is a vertical asymptote because the denominator tends to zero as tends to , while the numerator remains bounded away from zero as tends to . You can again make some intelligent change of variables to show that it comes down to the fact that has a vertical asymptote at if for all in some interval containing .

Okay cool thanks humph, so for Q 3, how would you do the same for the other question?

[the.watchman]

Well,

If you solve first, you get solutions or

as the denominator approaches 0, making y undefined.
From right, it approaches (numerator = 5, denominator is negative approaching 0).
From left, it approaches (numerator = 5, denominator is positive approaching 0).

Same with , I guess.
From right, it approaches (numerator = 1, denominator is positive approaching 0).
From left, it approaches (numerator = 1, denominator is negative approaching 0).

I'm not exactly that good at maths, so I could be really wrong

[TrueTears]

Well,

If you solve first, you get solutions or

as the denominator approaches 0, making y undefined.
From right, it approaches (numerator = 5, denominator is negative approaching 0).
From left, it approaches (numerator = 5, denominator is positive approaching 0).

Same with , I guess.
From right, it approaches (numerator = 1, denominator is positive approaching 0).
From left, it approaches (numerator = 1, denominator is negative approaching 0).

I'm not exactly that good at maths, so I could be really wrong

Yeah thanks, that is the way my book goes about 'explaining' it, but I was looking for a more rigorous 'proof' heh

Maybe I'll try a proof, but it seems too hard for this one =S

[TrueTears]

2. Also how to evaluate ? (How to set it out formally?)

Actually how would one use an proof on this?

We have to prove that for every there exists a number such that if then

Right?

[kamil9876]

Maybe I'll try a proof, but it seems too hard for this one =S

   
Contrary to first impressions, proofs are actually simply and neat if done for a more general result. When doing them on specific results like specific numbers and specific functions, certain textbooks can make the proofs messy. The same woudl happen for this one. But if you rather focus on proving limit laws first AND then apply them to this question, it is not so messy.

By request:

Suppose we want to find the as mentioned in TT's post.


We will take the following properties for granted:

Now we have that for every given , there exists an such that (this is the meaning of )

We also have that for every given (such as say that one given in the previous sentence) there exists a such that . (this is the meaning of .

Now applying this to the proof:

So in summary, we have that for every there exist and that give (ie )Which completes the proof.

Note: this is an existence proof, not a 'find an explicit expression for in terms of proof' that you are used to.

[TrueTears]

Maybe I'll try a proof, but it seems too hard for this one =S

   
Contrary to first impressions, proofs are actually simply and neat if done for a more general result. When doing them on specific results like specific numbers and specific functions, certain textbooks can make the proofs messy. The same woudl happen for this one. But if you rather focus on proving limit laws first AND then apply them to this question, it is not so messy.

By request:

Suppose we want to find the as mentioned in TT's post.


We will take the following properties for granted:

Now we have that for every given , there exists an such that (this is the meaning of )

We also have that for every given (such as say that one given in the previous sentence) there exists a such that . (this is the meaning of .

Now applying this to the proof:

So in summary, we have that for every there exist and that give (ie )Which completes the proof.

Note: this is an existence proof, not a 'find an explicit expression for in terms of proof' that you are used to.

omg that is so smart, thanks kamil.

[kamil9876]

By taking those properties as for granted, you can see that it can be extended to the general case of any continous functions that satisfy blah blah blah...

[TrueTears]

By taking those properties as for granted, you can see that it can be extended to the general case of any continous functions that satisfy blah blah blah...

Yeah, so it's just the conditions for

[TrueTears]

Also say we are using the precise definition to prove

This means that for every then th ere is a corresponding such that if then

Is it just simply

Thus if we have then the condition is satisfied?

[Damo17]

Also say we are using the precise definition to prove

This means that for every then th ere is a corresponding such that if then

Is it just simply

Thus if we have then the condition is satisfied?

Yes, that is correct.

[TrueTears]

Find and such that

Need to introduce something new here, but what lol

Thanks

[TrueTears]

Is there even any solutions, TT?  

yup

needs a bit of wishful thinking for this Q