Math Problem (Calculus)

[dekamii]

i know this is BG but worth asking. having a hard time solving this limit problem.

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

since x approaches infinity that means the x cancels out and we are left with +a/-a, no? how are we suppose to make +a/-a = e?

thanks!

[Blubbartron]

take the natural log of both sides. Don't forget ln(x^y) = y(ln x). Works out like butter from there.

[Pun]

Also, ln of e is 1.

[oldoldman]

Something seems off about that, when you are taking a limit there shouldn't be an equality. What that is saying is the limit of that formula is e, in which case there is nothing to solve.

[dekamii]

did you guys happen to get

a= (x-xe)/(1+e)?

that's what i got after taking the natural log of both sides.

@oldoldman:
i think the problem means that e is the value that you'll get as x approaches infinity. and the limit of that formula is indeed e but you have to find the value for a before it satisfies the =e.

[Blubbartron]

Ahh damn, oldoldman does seem to be correct. I wasn't thinking in my quick glance at it. Upon closer inspection, it might be an algebraically manipulated form of the actual definition for e. Is that what you're supposed to find? Beats me, now. Needs more context.

[oldoldman]

did you guys happen to get

a= (x-xe)/(1+e)?

that's what i got after taking the natural log of both sides.

@oldoldman:
i think the problem means that e is the value that you'll get as x approaches infinity. and the limit of that formula is indeed e but you have to find the value for a before it satisfies the =e.

The value for a is insignificant at infinity.

[Hosom]

What oldoldman said, something is off about it. Also, Check your work with this. WolframAlpha is basically the most amazing calculator you'll ever use. Don't have it do it for you, as you won't have it for a test, you'll just dig yourself into a hole that way.

[Teorem]

The value for a is insignificant at infinity.

When it's something simple like limit(x-a), x=>infinity, then sure, the value of a is not significant for very large values of x. However, that's not the case here because the function is a rational being raised to the power of x as well.

If you think about it, for values of a >= 0, (x+a)/(x-a) is always going to be >= 1, which means 1 and some remainder value.

If you know already that limit(1 + 1/x), x=> infinity = e, it's a matter of getting what you have into a similar expression. Think about how you can re-express the numerator to get that form.

Spoiler: show

x+a = (x-a) + 2a

[Ramor]

ln both sides and do some hocus-pocus magic with logarithmic rules and you'll end up with:

xln(x+a)-xln(x-a)=1 => x(ln(x+a)-ln(x-a))=1 => lim of bla bla bla ln(x+a)-ln(x-a)=1/x, 0=0?

of course this could very well be wrong, but i'm pretty sure oldoldman has it correct when he states that the value of a is insignificant at infinity, for instance:

lim[x->sideways8] of [(x+123081230912059123a)/(x-1095801289052390237348a)] where a=9999999999999999999999999999999999

is still going to be 1

[oldoldman]

When it's something simple like limit(x-a), x=>infinity, then sure, the value of a is not significant for very large values of x. However, that's not the case here because the function is a rational being raised to the power of x as well.

If you think about it, for values of a >= 0, (x+a)/(x-a) is always going to be >= 1, which means 1 and some remainder value.

If you know already that limit(1 + 1/x), x=> infinity = e, it's a matter of getting what you have into a similar expression. Think about how you can re-express the numerator to get that form.

Spoiler: show

x+a = (x-a) + 2a

If you are taking a "limit" of infinity you have to remember that you are actually just seeing what the end behavior of the variable would be, that is what it would be if there were no constants involved. Infinity is really hard to comprehend, I still have problems with it, but it basically is saying that no matter what you do to it, it's still infinite. The original problem was just a statement of the limit that converges to e, there is no problem.

[Ramor]

infinite, while simple to understand, is difficult to comprehend how actually immense it really is, i understand why a lot of people have trouble with it

in limits though you're entirely right, it just annuls (for most cases) any constant

[Teorem]

Again, the value of a is *not* insignificant in this case because the function is greater than one and being raised to a power of x. You cannot simply discard the constant value from the complete function. I agree that in cases like:

x+a
x-a
(x+a)/(x-a)

, then, yes, the value of a is not significant relative to x as x gets very large and approaches infinity. In the case of:

[(x+a)/(x-a)]^x

, the constant a has the effect of raising the bound of the limit from 1 to e. It's not an "insignificant" effect on the limit, even if you may consider the value of x >>>>>>>> a.

The original problem was just a statement of the limit that converges to e, there is no problem.

I'm aware of this, which is why I addressed it by trying to lead him to the equivalent and known expressions of e in limit form.

Edit:

In case I'm still not being clear... if you tell someone that "a is insignificant", when they look at the rational, they think "Oh, I can just remove a when x is really big, and then we have infinity divided by infinity which has to be one, which is still one when raised to any power, so the limit has to be one." Not much unlike other posts in this thread so far.

Which is wrong.

The value of a may be insignificant (very small in size, marginal) with respect to the value of x.
The value of a is not insignificant (ignorable) with respect to calculating the value of the limit.

[oldoldman]

Christ, you're right. Expanding the series makes it e^2a + ((2a^5*3e^2a)/(45x^4)) + ((2a^5(5a+9)(e^2a))/(45x^4) (ignoring the =e for now). Once you do that then you apply the limit, which would negate the last two terms and make the formula e^2a.
Now you put the equality and get e^2a=e, a=1/2.

[Daemeon]

How do you expand that equation? It doesn't resemble any of the common Taylor series.
@Ramor, you can't do ln(lim(a^x))=xln(lim(a)). But ln( (lim(a))^x )=xln(lim(a)), sure.

[oldoldman]

When I'm a bit more coherent tomorrow I might explain the expansion.

[Trajan]

i know this is BG but worth asking. having a hard time solving this limit problem.

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

since x approaches infinity that means the x cancels out and we are left with +a/-a, no? how are we suppose to make +a/-a = e?

thanks!

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

[cyaan]

think some ppl tried explaining, not sure if they are right or not. this should help though.

first step is making e to be raised to a power(e on its own is not a value, its a function). so we have (x+a)/(x-a)=e^(1/x) right?

ok now we have the limit as x is going to infinite, in which case the number "a" is so small compared to "x = infinite" that we can ignore it all together. so we have x/x=e^(1/x) as x approaches infinite.

lets make this look better: 1 = e ^(1/x), as x approaches infinite.

you should know that 1/infinite is 0, and anything raised to the 0 power is equal to 1. so 1 = e^(1/infinite) is a true statement.

at any rate, the limit of the function is 1.

edit
your kind of right, i did put some numbers backwards. ln(1)=0, and e^0=1. Also, e raised to the power of 1 is the number 2.7182, not e alone. much like "pi=3.14" people mistake a function for a number.

[Trajan]

think some ppl tried explaining, not sure if they are right or not. this should help though.

first step is making e to be raised to a power(e on its own is not a value, its a function). so we have (x+a)/(x-a)=e^(1/x) right?

ok now we have the limit as x is going to infinite, in which case the number "a" is so small compared to "x = infinite" that we can ignore it all together. so we have x/x=e^(1/x) as x approaches infinite.

lets make this look better: 1 = e ^(1/x), as x approaches infinite.

you should know that 1/infinite is 0, and anything raised to the 0 power is equal to 1. so 1 = e^(1/infinite) is a true statement.

at any rate, the limit of the function is 1.

you can check this with a calculator or graph it, as someone said ln(0)=1=e^0, or you can take some advice and get a solutions manual because everyone should have one.

There are so many things wrong with this statement it is not even funny. e IS a value (roughly 2.718281) because it is a real number and ln(0) is not equal to 1 it is undefined.

[cyaan]

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.