# Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}##

• Umrao
In summary: When I was posting the last thread, it was 5am here, and so I was feeling very sleepy, so I thought I am not getting it right now, I should try later. But even now that I have tried when I am wide awake I can't find the two products you mentioned.As in my last threadI am sorry but I could not pick up the clue. From your description I got.$$1/4 \lim_{y/2\to 0}[(\frac {e^{y/2}-y/2-1}{y/2})(\frac {e^{y/2}-y/2+1}{y/2}) Umrao ## Homework Statement Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}## without using L'Hospital's rule or expansion of the series. Answer is given to be = ##\frac{(\log_e (2))^2}{2}## ## Homework Equations Squeeze play theorem/ Sandwich theorem, some algebraic manipulations and standard results upto high school like$$\lim_{x\to 0}\frac{a^x-1}{x} = \log_e (a)$$Although we get the above formula by expansion but I think it would be valid to use it if we can. No I am not sure about it. These are all the the methods that I can think of. ## The Attempt at a Solution I have given multiple attempts to this question with a gap of several days in between, so I can't find all of them at this moment. I'll update the question when I find them. I have this one attempt with me at the time of posting this thread, so I'll write this down.$$\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}\rightarrow \lim_{x\to 0} \frac{e^{x\log_e 2}-1-x\log_e2}{x^2}$$putting ##x\log_e2 = y##$$\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-1-y}{y^2}\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1+ye^{y/2}-y}{y^2}\rightarrow (log_e 2)^2(\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2}+\lim_{y\to 0} \frac{ye^{y/2}-y}{y^2})$$Solving it part by part L1 =$$(log_e 2)^2(\lim_{y\to 0} \frac{ye^{y/2}-y}{y^2})\rightarrow (log_e 2)^2(\lim_{y\to 0} \frac{e^{y/2}-1}{2(y/2)}) = \frac{(log_e 2)^2}{2}$$L2 =$$(log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2}$$for ## 1 \geqslant y \geqslant -1##$$\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \geqslant (log_e 2)^2\lim_{y\to 0} e^{y}-ye^{y/2}-1\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \geqslant 0 $$That's it. I failed to find a function which can give ##\rightarrow (log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \leqslant 0 ## which would have completed the proof. Any help would be greatly appreciated. Also if anyone can find a function which binds the limit as ##(log_e 2)^2\lim_{y\to 0} \frac{e^{y}-ye^{y/2}-1}{y^2} \leqslant 0 ## , I would appreciate it if you can suggest some other ways than using sandwich theorem. mfb Have you tried using L'Hospital's rule? Note that you may have to use it more than once. Geofleur said: Have you tried using L'Hospital's rule? Note that you may have to use it more than once. The problem statement specifically precludes using L' Hopital's Rule. ...without using L'Hospital's rule ... Oh darn! Sorry about that.$$e^y - ye^{y/2} = (e^{y/2} - \frac{y}{2})^2 - \frac{y^2}{4}$$That gives the difference of two squares (not the two shown here), which you can convert into a product, and take care of the remainder separately. One term of this product has a trivial limit, the other one got rid of the nasty square in the denominator. Didn't check it in detail, but that looks working. Using y/2 instead of y as new variable looks more convenient. Umrao mfb said:$$e^y - ye^{y/2} = (e^{y/2} - \frac{y}{2})^2 - \frac{y^2}{4}$$That gives the difference of two squares (not the two shown here), which you can convert into a product, and take care of the remainder separately. One term of this product has a trivial limit, the other one got rid of the nasty square in the denominator. Didn't check it in detail, but that looks working. Using y/2 instead of y as new variable looks more convenient. I am sorry but I could not pick up the clue. From your description I got.$$1/4 \lim_{y/2\to 0}[(\frac {e^{y/2}-y/2-1}{y/2})(\frac {e^{y/2}-y/2+1}{y/2})-1]$$Am I going right? mfb said:$$e^y - ye^{y/2} = (e^{y/2} - \frac{y}{2})^2 - \frac{y^2}{4}$$That gives the difference of two squares (not the two shown here), which you can convert into a product, and take care of the remainder separately. One term of this product has a trivial limit, the other one got rid of the nasty square in the denominator. Didn't check it in detail, but that looks working. Using y/2 instead of y as new variable looks more convenient. When I was posting the last thread, it was 5am here, and so I was feeling very sleepy, so I thought I am not getting it right now, I should try later. But even now that I have tried when I am wide awake I can't find the two products you mentioned. As in my last thread Umrao said: I am sorry but I could not pick up the clue. From your description I got.$$1/4 \lim_{y/2\to 0}[(\frac {e^{y/2}-y/2-1}{y/2})(\frac {e^{y/2}-y/2+1}{y/2})-1]
Am I going right?

The two factors that I have are

## \frac {e^{y/2}-y/2-1}{y/2}## and ##\frac {e^{y/2}-y/2+1}{y/2}## but none of these appear to be trivial limit. Please tell me the products you were talking about.

Call ##f(x) = \frac{2^x -1 - x\ln 2}{x^2} ##.
You can write this under the form ## f(x) = \frac{h(x) - h(0)}{x} ##, where ##h(x) = \frac{2^x - 1}{x} ## if ## x \neq 0##, and ##h(0) = \ln 2##.
So if the limit in 0 exists, then ##f(x) \rightarrow_0 h'(0)##.
But for any ##x\neq 0##, ##h'(x) = -f(x) + h(x) \ln 2##, so by taking the limit as ##x## tends to 0 in both side, you get ## h'(0) = \frac{h(0) \ln 2}{2} = \frac{\ln^2 2}{2}##EDIT: sorry I think it is incorrect because I can't justify that ##h'(x) \rightarrow_0 h'(0)## on the left side of the equality

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Umrao
Sorry, you don't get rid of the square, that would make one factor divergent. Here is a slight modification: you can rearrange the denominators a bit: ##e^{y/2}-y/2+1## has a limit that is easy to evaluate. The other factor, after substitution, is a limit you encountered in a previous step.
I didn't check the prefactor in every step, but I think you get something like
(limit you want to know) = a*(limit you want to know) + c
with known a and c (and a not equal to 1). That's a linear equation you can solve.

Umrao
mfb said:
Sorry, you don't get rid of the square, that would make one factor divergent. Here is a slight modification: you can rearrange the denominators a bit: ##e^{y/2}-y/2+1## has a limit that is easy to evaluate. The other factor, after substitution, is a limit you encountered in a previous step.
I didn't check the prefactor in every step, but I think you get something like
(limit you want to know) = a*(limit you want to know) + c
with known a and c (and a not equal to 1). That's a linear equation you can solve.

Finally! Thanks a ton! Yipee! got the answer.
By the way, I was curious from your first post (on this thread) - what does - "mfb" stand for? Your initials?

Praise be unto He. Helium!

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mfb and Geofleur
Umrao said:
what does - "mfb" stand for? Your initials?
It had a meaning long ago that is no longer relevant, now it is just my nick.

Umrao
geoffrey159 said:
Call ##f(x) = \frac{2^x -1 - x\ln 2}{x^2} ##.
You can write this under the form ## f(x) = \frac{h(x) - h(0)}{x} ##, where ##h(x) = \frac{2^x - 1}{x} ## if ## x \neq 0##, and ##h(0) = \ln 2##.
So if the limit in 0 exists, then ##f(x) \rightarrow_0 h'(0)##.
But for any ##x\neq 0##, ##h'(x) = -f(x) + h(x) \ln 2##, so by taking the limit as ##x## tends to 0 in both side, you get ## h'(0) = \frac{h(0) \ln 2}{2} = \frac{\ln^2 2}{2}##EDIT: sorry I think it is incorrect because I can't justify that ##h'(x) \rightarrow_0 h'(0)## on the left side of the equality

Sorry for reopening the thread, but I've got an explanation why ##h'(x) \rightarrow_0 h'(0)##. I've re-read my notes about differentiation, and with this theorem, you can conclude :

Th: "Let ##x_0\in\mathbb{R}##, ##I ## an interval such that ##x_0\in \text{Int}(I)##, and ##f## a function from ##I## to ##\mathbb{R}##. If ##f## is continuous in ##x_0##, differentiable on ##I-\{x_0\}##, and ##f'(x)## tends toward a limit ##\ell## in ##x_0##, then ##f## is differentiable in ##x_0## and ## f'(x_0) = \ell ##"

Here ##h## is continuous in 0, differentiable on ##\mathbb{R} -\{0\}## such that ##h'(x) = -f(x) + h(x)\ln 2 ##, ##h'(x)\rightarrow_0 -h'(0) + h(0)\ln 2## which is finite with the assumption ##f## converges in 0. Therefore ## h'(0) = -h'(0) + h(0)\ln 2##, and the only possible limit is ##\frac{\ln^2 2}{2}##

geoffrey159 said:
Sorry for reopening the thread, but I've got an explanation why ##h'(x) \rightarrow_0 h'(0)##. I've re-read my notes about differentiation, and with this theorem, you can conclude :

Th: "Let ##x_0\in\mathbb{R}##, ##I ## an interval such that ##x_0\in \text{Int}(I)##, and ##f## a function from ##I## to ##\mathbb{R}##. If ##f## is continuous in ##x_0##, differentiable on ##I-\{x_0\}##, and ##f'(x)## tends toward a limit ##\ell## in ##x_0##, then ##f## is differentiable in ##x_0## and ## f'(x_0) = \ell ##"

Here ##h## is continuous in 0, differentiable on ##\mathbb{R} -\{0\}## such that ##h'(x) = -f(x) + h(x)\ln 2 ##, ##h'(x)\rightarrow_0 -h'(0) + h(0)\ln 2## which is finite with the assumption ##f## converges in 0. Therefore ## h'(0) = -h'(0) + h(0)\ln 2##, and the only possible limit is ##\frac{\ln^2 2}{2}##
I think h is not continuous at 0. limit h(0) does exist but limit has to be equal to h(0) which is undefined.
h(0) = infinity/infinty. You had h(0) = ln2 by taking the limit (using a standard formula.)

Umrao said:
I think h is not continuous at 0. limit h(0) does exist but limit has to be equal to h(0) which is undefined.

It is continuous because ##h(x) \rightarrow_0 \frac{d}{dx}(2^x)(0) = \ln 2 = h(0)##

geoffrey159 said:
It is continuous because ##h(x) \rightarrow_0 \frac{d}{dx}(2^x)(0) = \ln 2 = h(0)##

I didn't put much thought to what was implied by differentiation (sorry I'll try to figure it out.) My book says:
"A function f(x) is said to be continuous at x = a; where a is an element of domain of f(x).
If left hand limit at a = right hand limit at a = f(a)
i.e. LHL=RHL=Value of function at x=a......
...
If f(x) is discontinuous at x =a ...
(I) either LHL is not equal to RHL
(II) LHL and RHL exist and are not equal to f(a)
(III) f(a) does not exist.
(IV) At least one of the limit does not exist."

I specifically remember an example where limits existed but the function was discontinuous, think about a function having limit at x = a as infinity, limits exist but function is discontinuous at a. I know that here we have ln2, think about this---> What if it is a point discontinuity (which exactly I think it is.)

geoffrey159 said:
It is continuous because ##h(x) \rightarrow_0 \frac{d}{dx}(2^x)(0) = \ln 2 = h(0)##

Cross checked with wolfram alpha, it says that h(x) is discontinuous at x=0.

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What is ##\lim_0 h(x)## equal to ?
What is ##h(0)## equal to ?
Do you see they are equal ?
This equality means (rigorously) that ##h## is continuous in 0.

h(x) is continuous, but I think you just reinvented a special case of l'Hospital.

## 1. What does this limit represent?

This limit represents the instantaneous rate of change of the function f(x) = 2^x at x = 0.

## 2. How do I evaluate this limit?

To evaluate this limit, you can use L'Hospital's rule or the power series expansion for 2^x.

## 3. What is L'Hospital's rule?

L'Hospital's rule states that if the limit of the ratio of two functions f(x) and g(x) as x approaches a certain value a is indeterminate, then the limit of the ratio of their derivatives f'(x) and g'(x) as x approaches a will be equal to the same limit of the original ratio.

## 4. How do I use the power series expansion to evaluate this limit?

The power series expansion for 2^x is 1 + x\log_e2 + \frac{x^2 \log_e^2 2}{2!} + \frac{x^3 \log_e^3 2}{3!} + .... You can plug this into the original limit and simplify to evaluate the limit.

## 5. What is the value of this limit?

The value of this limit is \log_e^2 2, or approximately 0.480453.

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