Math Problem (Calculus)

[Xajii]

You might be having a hard time solving this because it isn't true. Try putting in some values for a. For a = 1 you find the limit to be 7.3891 or e^2. For a = 3 you get 403.42 or e^6.

Your problem should look like this: Lim x->inf ((x+a)/(x-a))^x = e^2a

This is correct

If you are having trouble understanding this

Instead of infinity just use 100,000

x = 100,000
a = 3

100,000 +3 / 100,000 -3 =

100,003 / 99,997 = 1.00006

1.00006 ^ 100,000 = 403.42

e^6 = 403.42

good work man @_@

[Trajan]

thanks for catching that, i edited my post.

also if you would like to clarify the "so many things" wrong with my post besides the typo at the end of it, and your misinformed view of the exponential function e, id be glad to defend those too.

My misinformed view? Let us be precise here, e, Euler's number, is a REAL NUMBER, you can find it on the REAL NUMBER LINE. All the exponential function, e^x, does is take different powers of the real number e. I.e. e^3 = (2.7182)^3 = 20.08.

Edit: All I am trying to say here is that there is a difference between the mathematical constant e, which has a precise value, and the exponential function e^x.

[Purrrfect]

It might be one of those problems where you are supposed to identify whether the numerator or denominator is the more controlling influence on the graph? Maybe looking for an asymptote?

[cyaan]

yeah im not saying it doesnt hold value, but i view it as a function.

we find precise values for such numbers through functions, so the number equals the function(my calculus book i just took out the closet defines e as the number such that the natural log of e is equal to 1 [e is a function of ln, also x] >> this is how we find it to be 2.71 and the decimals can keep going if you expand on the series) so what i am saying is the precise value of e, and the function e^x, are the same.

im just trying to help the kid out not argue/debate. been a while since i did calculus let alone at 1am so if i was wrong in my solution i appoligize.

to the OP get a solutions manual if you plan on going further into calculus, and the web isnt a substitute for real ppl helping you in real time. This is my real, sincere advice.

[Trajan]

I was bored this morning so I figured out some generalizations if anyone is interested.

Lim x -> inf ((x+a)/(x-b))^x = e^(a+b) and Lim x-> inf ((x-a)/(x+b))^x = e^-(a+b)

[dekamii]

thanks for the insights, guys! this is one of the extra credit problems which by no means suppose to be easy. i guess i'll just have to talk to the professor. thanks again!

[Trajan]

thanks for the insights, guys! this is one of the extra credit problems which by no means suppose to be easy. i guess i'll just have to talk to the professor. thanks again!

If you are interested in the solution to the problem I solved it earlier in the day, it is not easy at all and if it wasn't for the website that Hosom had linked on his post I probably would not have solved it b/c one of the derivatives is really nasty.

[Galois]

My approach would be to attempt to get the limit of the form lim{x-> inf} (1+(1/x))^x, which is a representation of e.

I've come to a block, but I am on the order of the following:

lim{x->inf} [(x+a)/(x-a)]^x
=lim{x->inf} [((x-a)+2a)/(x-a)]^x
=lim{x->inf} [(x-a)/(x-a) + (2a/(x-a))]^x
=lim{x->inf} [1+ (2a/(x-a))]^x

From there I'd wager that there's a clever trick to show that expression is equivalent to the above representation of e.

[CDF]

I don't think anyone explicitly stated this, but the real "trick" here, as shown in the previous post, is to rewrite (x+a)/(x-a) as 1 + 2a/(x-a). Afterwards, take the log of the expression and find the limit of this log-transformed expression using l'Hopital's rule twice. You will get 2a, then exponentiate to get the final result, exp(2a).

I'm guessing the point of this problem was to get people to think about how to rewrite the expression so that you can use l'Hopital's rule.

[Galois]

I also noticed something in earlier posts that I meant to mention in my last post. Some instructors can be strict over using the result of the problem to solve the problem. What I mean is, using that fact that the limit of the expression is e to show that the limit is e.

I remember an instructor that would not allow students to do such things while working on trigonometric identities, such as multiplying both sides by sin(x) to clear out the denominators while trying to show an identity.

Like I said, some instructors are more strict and some are lenient with such rules, but I've found that more fall into the former category rather than the latter.

[Leif]

What I mean is, using that fact that the limit of the expression is e to show that the limit is e.

...

If nobody has got it yet, the whole purpose of the problem is to prove that is equal to e, not so you can use some fake math step to plug in the original problem which may or may not be accurate in order to prove something. The answer is in the work you do to prove that the limit is e.

[Galois]

Perhaps you misunderstood my post (or maybe I am misunderstanding yours).

My point is that, in general, one cannot assume what they are proving to be true in order to show it is true.

For example, if one wanted to show that the trigonometric identity

(sec x)^2 + (csc x)^2 = (sec x)^2 * (csc x)^2

was true, one does not want to multiply each side by the expression

(cos x)^2 * (sin x)^2

since by doing so, one is assuming that the equality is already true. In my experience, though, I have come across some instructors that are not as opposed to that step as others.

[Trajan]

I don't think anyone explicitly stated this, but the real "trick" here, as shown in the previous post, is to rewrite (x+a)/(x-a) as 1 + 2a/(x-a). Afterwards, take the log of the expression and find the limit of this log-transformed expression using l'Hopital's rule twice. You will get 2a, then exponentiate to get the final result, exp(2a).

I'm guessing the point of this problem was to get people to think about how to rewrite the expression so that you can use l'Hopital's rule.

Almost, but you really don't need to rewrite it like that. This is the solution to the problem:

You can't assume that the limit is equal to e^2a right off the bat, so set it equal to some new variable z (and we show that z must equal e^2a). You start off with Lim x -> inf ((x+a)/(x-a))^x = z.

Next, use the fact that Ln( Lim x-> inf f(x)^(g(x))) = Lim x -> inf Ln(f(x))/(1/g(x)) (I had to look this up). In this form you can use L'hospitals rule. So you set f(x) = (x+a)/(x-a) and g(x) = x.

Note that you have to take the Ln of both sides, so you now have:
Lim x -> inf ln(((x+a)/(x-a))/(1/x)) = ln(z). You can now use L'hospital's rule. I used the website that Hosom had linked b/c the derivative of ln((x+a)/(x-a)) is nasty.

After you have taken the derivatives and simplified you should end up with this:
Lim x -> inf (2ax^2)/(x^2-a^2) = ln(z). As x -> inf x^2/(x^2-a^2) -> 1.

Finally, you are left with 2a = ln(z), so z = e^2a, and you are done (whew!).

[CDF]

If nobody has got it yet, the whole purpose of the problem is to prove that is equal to e, not so you can use some fake math step to plug in the original problem which may or may not be accurate in order to prove something. The answer is in the work you do to prove that the limit is e.

No. The point is to use concepts learned from differential calculus to evaluate a limit, which is not even e.

Almost, but you really don't need to rewrite it like that. This is the solution to the problem:

You can't assume that the limit is equal to e^2a right off the bat, so set it equal to some new variable z (and we show that z must equal e^2a). You start off with Lim x -> inf ((x+a)/(x-a))^x = z.

Next, use the fact that Ln( Lim x-> inf f(x)^(g(x))) = Lim x -> inf Ln(f(x))/(1/g(x)) (I had to look this up). In this form you can use L'hospitals rule. So you set f(x) = (x+a)/(x-a) and g(x) = x.

Note that you have to take the Ln of both sides, so you now have:
Lim x -> inf ln(((x+a)/(x-a))/(1/x)) = ln(z). You can now use L'hospital's rule. I used the website that Hosom had linked b/c the derivative of ln((x+a)/(x-a)) is nasty.

After you have taken the derivatives and simplified you should end up with this:
Lim x -> inf (2ax^2)/(x^2-a^2) = ln(z). As x -> inf x^2/(x^2-a^2) -> 1.

Finally, you are left with 2a = ln(z), so z = e^2a, and you are done (whew!).

Good job spending many more words to describe the same thing I did (are we holding people's hands?) while emphasizing that the derivative of ln (x+a)/(x-a) is somehow "nasty."

[Trajan]

No. The point is to use concepts learned from differential calculus to evaluate a limit, which is not even e.



Good job spending many more words to describe the same thing I did (are we holding people's hands?) while emphasizing that the derivative of ln (x+a)/(x-a) is somehow "nasty."

You were right in your original post that the point of the problem was to get it into the form to be able to use L'hospitals rule. But there isn't any "trick" being used here, you don't need to rewrite it into the form of 1 + 2a/(x-a) and you don't need to use L'hospitals rule twice, only once. And yes the derivative of ln((x+a)/(x-a)) is hard for me now that I have been out of college and haven't had a math class in 2 years.

[CDF]

Based on the kinds of non-responses in this thread, the difficulty for some here was not using the tactic of log-transforming, but recognizing that ln[(x+a)/(x-a)] tends to 0 in the limit.

Recognizing it is one thing, but justifying it (showing work) is another. What I called a "trick" is kind of akin to manipulation techniques like completing the square in the sense that it can help make it easier to see ln[(x+a)/(x-a)] is really 0 in the limit.

Finally, taking the limit of x^2/(x^2-a^2) is justified by l'Hopital's rule anyway (again recognizing and justifying are different), so strictly speaking l'Hopital's rule is "used" twice whether implied or not.

[Zanmato]

There are so many things that are not right I’m not too sure where to start.

having a hard time solving this limit problem.

lim(x->infinity) ((x+a)/(x-a))^x = e

The problem here is that there isn’t really any problem. It has an “equal to” sign and the answer (e) as well so I’m not too sure which of the following you’re asking.

- Do you want to prove lim(x->infinity) ((x+a)/(x-a))^x equals e?
- Or are you asking what is the limit of ((x+a)/(x-a))^x as (x->infinity)?
- Or are you asking for what value of a will lim(x->infinity) ((x+a)/(x-a))^x = e? If this is the case, the solution you’re looking for is in Post #38. a = 1/2.

After doing some math, I have concluded that the limit does not always equal to e. Therefore “lim(x->infinity) ((x+a)/(x-a))^x = e” is correct, but to some extent.

If you plug in x = 100, a = 0.5, you get 2.7183044 which is pretty close to e. If you plug in x = 1000, a = 0.5, you get 2.718282 which is even closer to e.

The thing is once u start playing with the value of a, lim(x->infinity) ((x+a)/(x-a))^x doesn’t really equal to e anymore. I don't know (yet) what it equals, but point is it doesn’t equal to e, which is the opposite of what you’re asking us to show (assuming you want us to show that it equals e).

I will try and figure out a hand written solution for lim(x->infinity) ((x+a)/(x-a))^x

[Zanmato]

Here is the hand written solution. I know it’s not a good thing to just give out the solution so I’ll explain each step as much as I can so you can understand better. Don’t flame me for writing an essay instead of solving a Math problem, I’m just trying to help.

This is almost identical to someone else’s solution and I admit reading his/her solution before coming up with this so inb4 “you stole my solution”. Again, it is pretty much what has already been said in a previous post, but without the use of WolframAlpha (website that Hosom linked to calculate derivatives).

Let lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x = z
So now in other words, we need to calculate the value of z.

ln [ lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x ] = ln [ z ]
Add ln to both sides. This is do-able since ln is an increasing function. (I don’t know about you guys but during my undergraduate career at University, I was required to state the “since ln is an increasing function” part for full marks so you might want to state that).

lim { x -> infinity } ln [ ( ( x + a ) / ( x – a ) ) ^ x ] = ln [ z ]
Similar logic to “the limit of a product is the product of the limits” so you can move the ln inside.

lim { x -> infinity } x ln [ ( ( x + a ) / ( x – a ) ) ] = ln [ z ]
All I did was bring down the power to the front. There is a logarithmic property that states ln [ m ^ n ] = n ln [ m ] where in this case:
n = x
m = ( x + a ) / ( x – a )

Here comes the tricky part. It is not really how smart you are to be able to see this but rather how experienced you are to be able to notice that this can be written in a form with which l'Hôpital's rule can be applied. (My Prof at University once said in Math, “Practice makes perfect, you can’t read and learn Math”).

lim { x -> infinity } ( ln [ ( ( x + a ) / ( x – a ) ) ] ) / ( 1 / x ) = ln [ z ]
All I did here is write it differently. Something like that; n ln [ m ] = ln [ m ] / ( 1 / n ). Since this limit as { x -> infinity } is in the form of “infinity divided by infinity” l'Hôpital's rule can be used. You can’t just differentiate and say you used l'Hôpital's rule, you have to state why l'Hôpital's rule can be used.

You now have to differentiate:
F ( x ) = ln [ ( ( x + a ) / ( x – a ) ) ]
And
G ( x ) = ( 1 / x )

First do some simplifying to differentiate:
F ( x ) = ln [ ( ( x + a ) / ( x – a ) ) ]
= ln [ x + a ] – ln [ x – a ] since ln [ p / q ] = ln [ p ] – ln [ q ].
So, F ‘ ( x ) = f ( x ) = ( 1 / ( x + a ) ) – ( 1 / ( x – a ) ) = –2a / ( x^2 – a^2 )
And G ‘ ( x ) = g ( x ) = –1 / ( x^2 ).

So basically:
ln [ lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x ]
= lim { x -> infinity } [ –2a / ( x^2 – a^2 ) ] / [ –1 / ( x^2 ) ]
Simplify and we get:
= lim { x -> infinity } [ 2a(x^2) / ( x^2 – a^2 ) ]
Gets easy from here:
= lim { x -> infinity } [ 2a ] / [ 1 – ( ( a^2 ) / ( x^2 ) ) ]
Well now this is what happens when { x -> infinity } or basically plug in x = infinity:
= 2a / ( 1 – 0 )
= 2a

In conclusion:

ln [ lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x ] = 2a

Therefore:

lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x = e^(2a)
Once again, you need to state this step can be done since e^x is an increasing function.

Done.

TAAAAAADDDDDAAAAAAAAAA!!

Now if your original question was for what value of a will

lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x

equal to e, then all you have to do (after the math I just did) is say:

Since lim { x -> infinity } ( ( x + a ) / ( x – a ) ) ^ x = e^(2a)
Set e^(2a) = e
Therefore 2a = 1 and so a = 0.5

MOAR TAAAAAAAADDDDDDDAAAAAAAAAAAA!!

I'm just wondering, what level of Math is this? High school, University, College?
Is there anyway you could post the solution your prof gave you for this?

[Trajan]

You stole my solution!!! Lol jk. Very nice.

[dekamii]

it's from Calculus II in university. the prof said he won't give out the solution until we hand it in. and thanks again for the solution guys!