Math question

[Kerr]

Hi all.

Have a BIG test tomorrow and I am puzzed at a math thing. Why is 2 the limit of x*(e^(2/x)-1)?

At least thats what mathematica says and I want to understand why. I mean, I have noticed a pattern.
x*(e^(2/x)-1) -> 2 when x->infinity
x*(e^(3/x)-1)-> 3 when x->infinity
x*(e^(4/x)-1) -> 4 when x->infinity
x*(e^(5/x)-1) -> 5 when x->infinity
...
x*(e^(2382/x)-1) -> 2382 when x->infinity

It seems like x*(e^(a/x)-1)-> a when x->infinity.

Why?

Take care,
Kerr.

 

[LuisDantas]

Allow me to ask if this is correct. You are saying that

and more generally,

and you want to know why that is so. Is that right? Not trying to be a smart aleck. I just happened to have a LaTex tool at hand and I find that it can be a real help in such situations.

I do not actually know the answer, nor whether the premise is correct. I will attempt to find out, though. No promises, I fear.

 

[LuisDantas]

Okay, here is what I found so far.

According to The Number e,

I believe you can go from there to

although I do not yet know how.

 

[Kerr]

Allow me to ask if this is correct. You are saying that

and more generally,

and you want to know why that is so. Is that right? Not trying to be a smart aleck. I just happened to have a LaTex tool at hand and I find that it can be a real help in such situations.

I do not actually know the answer, nor whether the premise is correct. I will attempt to find out, though. No promises, I fear.

Lol, yeah, I wish I knew how to write it in a more mathematical way . But you are correct. I did look it up, and you can solve it by setting t = a / x and then do this:

lim[x->Infinity] x*(e^(a/x)-1) = lim[t->0] a*(e^t-1)/t

And then you can use a standard limit thing that says that (e^x-1)/x -> 1 when x -> 0. But that doesnt make me understand why the first one didnt work. So, basically, now I wish to know why I needed to make a variable substitution. Because when I look at the original one I get:

lim[x->Infinity] x*(e^(a/x)-1) = Infinity

Or something. I dont know. Its odd because (e^(a/x)-1) -> 0 while x ->infinity, and I am not sure what to make of that. Probably I could make it easier for myself by just trying to learn the standard limit things and apply them without thought, but my brain wants to understand .

 

[Kerr]

Okay, here is what I found so far.

According to The Number e,

I believe you can go from there to

although I do not yet know how.

Yeah, it might have something to do with it. I just dont have the time to explore it. As I said, major test tomorrow, its better to fail at many then succeed at one because then I have gotten more exposure to math .

 

[WonderWho]

Let A be any real value.
Let S=x*(e^(A/x)-1), then since
e^y= 1+y + y²/2! + y³/3!..+y^n/n! + ....

S= x*((A/x) + (A/x)A/2! + (A/x)³/3!..+(A/x)^n/n! + ....
= x*((A/x) + (A/x)²/2! + (A/x)³/3!..+ ....
= A + (A²/2x) +(A³/6x²) + ...

as x->∞, all the terms on the RHS tend to zero except the first term, leaving A as the limit.

 

[Kerr]

Let A be any real value.
Let S=x*(e^(A/x)-1), then since
e^y= 1+y + y²/2! + y³/3!..+y^n/n! + ....

S= x*((A/x) + (A/x)A/2! + (A/x)³/3!..+(A/x)^n/n! + ....
= x*((A/x) + (A/x)²/2! + (A/x)³/3!..+ ....
= A + (A²/2x) +(A³/6x²) + ...

as x->∞, all the terms on the RHS tend to zero except the first term, leaving A as the limit.

Thanks . Will take a closer look when I get the time.

 

[LegionOnomaMoi]

then since
e^y= 1+y + y²/2! + y³/3!..+y^n/n! + ....

This is the series expansion if y=0. In general,

S= x*((A/x) + (A/x)A/2! + (A/x)³/3!..+(A/x)^n/n! + ....
= x*((A/x) + (A/x)²/2! + (A/x)³/3!..+ ....
= A + (A²/2x) +(A³/6x²) + ...

as x->∞, all the terms on the RHS tend to zero except the first term, leaving A as the limit.

I agree that the obvious thing to do is use the properties of e here and factor out the constant, but I'm not sure (it's hard to read the series notation in that form) if the above is true or whether it assumes y=0 without getting us to a point where we can make y=0. Either way, it may be simpler to substitute another function in for y that allows us to use L'Hopital's.

Let x= 1/k. Then the limit becomes

Now we have an indeterminate form of the type lim as x-> infinity f(x) / g(x) and we can apply L'Hopital's and get

Factor out the constant

Now we can make "y" (i.e., k) equal to 0 by factoring out the 2 in the exponent:

This leaves us with the constant you gave above (which, in this case, is 2).