Limits: show that the following limit doesn't depend on the election of "b".

[elesar123]

Hi, have this problem, can't solve it.
Given "a" and "b" real numers, I have to show that the following limit doesn't depend on the election of "b".
VERY IMPORTANT: I can't use L'Hoppital's rule in any way or derivatives, since it's not supposed to be part of this section of the course (those things are posterior, so I should have to solve the limit without it).
Essentialy I'm asking how to solve the limit just by transforming the expression, I guess. Thanks

Lim (x-->+inf) of [1 + (a/x) + (b/x^2)]^x

 

[JeffM]

What is [imath]\displaystyle \lim_{x \rightarrow \infty} 1^x[/imath]?

 

[blamocur]

Hi, have this problem, can't solve it.
Given "a" and "b" real numers, I have to show that the following limit doesn't depend on the election of "b".
VERY IMPORTANT: I can't use L'Hoppital's rule in any way or derivatives, since it's not supposed to be part of this section of the course (those things are posterior, so I should have to solve the limit without it).
Essentialy I'm asking how to solve the limit just by transforming the expression, I guess. Thanks

Lim (x-->+inf) of [1 + (a/x) + (b/x^2)]^x

Can you figure out [imath]\lim_{x->\infty} \left(1 + \frac{b}{x^2}\right)^x[/imath]?

 

[lookagain]

What is [imath]\displaystyle \lim_{x \rightarrow \infty} 1^x[/imath]

This post should not be given any thumbs-up, because the problem does not involve
the limit of 1^x as x approaches infinity.

The problem does involve the indeterminate form \(\displaystyle \ 1^{oo}\).

 

[lookagain]

VERY IMPORTANT: I can't use L'Hoppital's rule in any way or derivatives, ...

Lim (x-->+inf) of [1 + (a/x) + (b/x^2)]^x

You have got to be able to use a tool such as the limit
as x ---> oo of (1 + u/x)^x = e^u for appropriate u.


A method:

Multiply \(\displaystyle \ \dfrac{b}{x^2} \ \) by 1/x on both its numerator and its denominator to obtain:

\(\displaystyle \dfrac{( \tfrac{b}{x})}{x}\)

Add this to a/x and 1. Next, put a/x and the above
fraction over the common denominator of x, with 1
added off to the left:

\(\displaystyle \bigg(1 + \dfrac{ \ a + b/x}{x}\bigg)^x\)

All the while there is intended to be the limit as x ---> oo
in front of these expressions.

The application of the last leads to
lim (x ---> oo) of e^(a + b/x).

What does this work out to be?

 

[elesar123]

You have got to be able to use a tool such as the limit
as x ---> oo of (1 + u/x)^x = e^u for appropriate u.


A method:

Multiply \(\displaystyle \ \dfrac{b}{x^2} \ \) by 1/x on both its numerator and its denominator to obtain:

\(\displaystyle \dfrac{( \tfrac{b}{x})}{x}\)

Add this to a/x and 1. Next, put a/x and the above
fraction over the common denominator of x, with 1
added off to the left:

\(\displaystyle \bigg(1 + \dfrac{ \ a + b/x}{x}\bigg)^x\)

All the while there is intended to be the limit as x ---> oo
in front of these expressions.

The application of the last leads to
lim (x ---> oo) of e^(a + b/x).

What does this work out to be?

I had actually reached that point but didn't realize that the final expression e^(a + b/x) was gonna be reduced to e^a beacase b/x when (x ---> oo) is zero. Detailed and amazing answer, Thank you very much!!!!

 

[lookagain]

I had actually reached that point but didn't realize that the final expression e^(a + b/x) was gonna be reduced to e^a beacase b/x when (x ---> oo) is cero. Detailed and amazing answer, Thank you very much!!!!

It was not my intent to provide as many steps as I did. I typed on a cell phone.
We want you to be able to contribute toward your solution as well.

 

[Steven G]

What is [imath]\displaystyle \lim_{x \rightarrow \infty} 1^x[/imath]?

Jeff, that 1 in 1^x is not a true 1. Sure [imath]\displaystyle \lim_{x \rightarrow \infty} 1^x = 1[/imath], but that is not what was given.
By a true 1, I mean the base is 1, not approaching 1

 

[JeffM]

Jeff, that 1 in 1^x is not a true 1. Sure [imath]\displaystyle \lim_{x \rightarrow \infty} 1^x = 1[/imath], but that is not what was given.
By a true 1, I mean the base is 1, not approaching 1

Jomo,

I get your distinction between true 1 and a limit of 1: the argument never equals the limit. I am, however, too tired to discuss. But I want to think whether the limit of a well defined limit is not equal to that limit. That is

[math]\lim_{x \rightarrow \infty} \left (1 + \dfrac{a}{x} + \dfrac{b}{x^2} \right ) = 1 \text { and } \lim_{x \rightarrow \infty} 1^x = 1.[/math]
I will ponder why the intuitive conclusion is false. Not disputing, just not grasping.