# A limit problem without the use of a Taylor series expansion

#### Physics lover

Homework Statement
This is a famous problem of limit.
Lim ((1+x)^(1/x)-e+ex/2)/x^2 where x tends to 0.

We can easily solve it with taylor expansion but i want to solve it without using taylor expansion.Is there a way?
Homework Equations
Taylor expansion
I tried substituting x=cos2theta but it was of no use.I thought many ways but i could not make 0/0 form.So please help.

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#### fresh_42

Mentor
2018 Award
Can you type this in LaTeX so that the function can be read more easily? See https://www.physicsforums.com/help/latexhelp/ for how it is done. I have read $\lim_{x\to 0} \dfrac{(1+x)^{\frac{1}{x}}-e-e\cdot \frac{x}{2}}{x^2}$ but I'm not sure. Have you tried L'Hôpital?

#### Physics lover

Can you type this in LaTeX so that the function can be read more easily? See https://www.physicsforums.com/help/latexhelp/ for how it is done. I have read $\lim_{x\to 0} \dfrac{(1+x)^{\frac{1}{x}}-e-e\cdot \frac{x}{2}}{x^2}$ but I'm not sure. Have you tried L'Hôpital?
Yes you have taken it correct.I think we cannot apply L' Hospital since 1^infinity is there and therefore no 0/0 or infinity/infinity form.

#### fresh_42

Mentor
2018 Award
And why isn't it simply $-\infty$?
$\lim_{x\to 0} \dfrac{(1+x)^{\frac{1}{x}}-e-e\cdot \frac{x}{2}}{x^2}= \lim_{x\to 0} \dfrac{e-e-e\cdot \frac{x}{2}}{x^2} = \lim_{x\to 0} \dfrac{-e}{2x}$

#### Physics lover

And why isn't it simply $-\infty$?
$\lim_{x\to 0} \dfrac{(1+x)^{\frac{1}{x}}-e-e\cdot \frac{x}{2}}{x^2}= \lim_{x\to 0} \dfrac{e-e-e\cdot \frac{x}{2}}{x^2} = \lim_{x\to 0} \dfrac{-e}{2x}$
So will L'Hospital work here.

#### fresh_42

Mentor
2018 Award
I haven't used L'Hôpital, only that $\lim_{x \to 0} (x+1)^{\frac{1}{x}} =e$. I think that is a bit cheating, as I ignored the denominator, so it has to be shown formally that $\lim_{x\to 0} \dfrac{(x+1)^{\frac{1}{x}}-e}{x^2}=0$ by whatever method.

#### Physics lover

I tried L'Hospital but it is becoming complicated.

#### Physics lover

I haven't used L'Hôpital, only that $\lim_{x \to 0} (x+1)^{\frac{1}{x}} =e$. I think that is a bit cheating, as I ignored the denominator, so it has to be shown formally that $\lim_{x\to 0} \dfrac{(x+1)^{\frac{1}{x}}-e}{x^2}=0$ by whatever method.
The answer is -11e/24.

#### fresh_42

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2018 Award
The answer is -11e/24.
Not according to WolframAlpha:

#### Physics lover

Not according to WolframAlpha:
But answer to a limit is never taken infinity and also by going by taylor expansion i get the answer as -11e/24.

#### fresh_42

Mentor
2018 Award
This is because I made a sign error in post #2 which you confirmed in post #3. Please use LaTeX next time. Seems we cannot simplify $(1+x)^{1/x}=e$ in the limit. I had such a feeling. In this case I would try L'Hôpital, but Taylor is easier.

#### Physics lover

This is because I made a sign error in post #2 which you confirmed in post #3. Please use LaTeX next time. Seems we cannot simplify $(1+x)^{1/x}=e$ in the limit. I had such a feeling. In this case I would try L'Hôpital, but Taylor is easier.
Ok i will use LaTeX next time.But can you please solve it without Taylor.

#### fresh_42

Mentor
2018 Award
Ok i will use LaTeX next time.But can you please solve it without Taylor.
This means in other words: Can you prove that the third term of the Taylor series of $\dfrac{(1+x)^\frac{1}{x}}{x^2}$ equals $\frac{11}{24}\,e$ without using the Taylor series? However, already $f''(0)$ is an ugly limit. Another wording is: How fast exactly does $(1+x)^\frac{1}{x}$ converge to $e$.

I'm sure there is a nice trick to manage those expressions, but I don't know any.

#### Physics lover

This means in other words: Can you prove that the third term of the Taylor series of $\dfrac{(1+x)^\frac{1}{x}}{x^2}$ equals $\frac{11}{24}\,e$ without using the Taylor series? However, already $f''(0)$ is an ugly limit. Another wording is: How fast exactly does $(1+x)^\frac{1}{x}$ converge to $e$.

I'm sure there is a nice trick to manage those expressions, but I don't know any.

Homework Helper
The limit is not -11e/24 -- the limit does not exist.

a) You cannot expand the numerator in a series around 0 since the exponential term is not defined for x = 0, so that approach does not apply

b) More directly, a plot of the expression (look at the plot from (say) x = 0.001 to about x = 0.3) shows a vertical asymptote tending to negative infinity as you approach 0 from the right.

#### fresh_42

Mentor
2018 Award
The limit is not -11e/24 -- the limit does not exist.

a) You cannot expand the numerator in a series around 0 since the exponential term is not defined for x = 0, so that approach does not apply

b) More directly, a plot of the expression (look at the plot from (say) x = 0.001 to about x = 0.3) shows a vertical asymptote tending to negative infinity as you approach 0 from the right.
WolframAlpha says differently:

Homework Helper
I find Alpha's "series expansion" suspicious. The requirement for that to be valid are that the function in question be differentiable about the point in question (0, in this case), and this function is not defined at 0.

Also: it seems that there are two expressions being bantered about here. The original expression was

$\lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e + \frac{ex}2}{x^2}$

The other is
$\lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e - \frac{ex}2}{x^2}$

Which one is correct?

#### fresh_42

Mentor
2018 Award
I find Alpha's "series expansion" suspicious. The requirement for that to be valid are that the function in question be differentiable about the point in question (0, in this case), and this function is not defined at 0.

Also: it seems that there are two expressions being bantered about here. The original expression was

$\lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e + \frac{ex}2}{x^2}$

The other is
$\lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e - \frac{ex}2}{x^2}$

Which one is correct?
See post #11.

#### fresh_42

Mentor
2018 Award
No, I have no idea. I tried a few things but they led nowhere. The margin is very, very narrow and one will need many terms of the Taylor expansion of $(1+x)^{1/x}$. I substituted $n= 1/x$ and worked with natural numbers, but calculation with just $O(\frac{1}{n})$ was not precise enough. Even integration uses the series expansion.

Last edited:

#### SammyS

Staff Emeritus
Homework Helper
Gold Member
I find Alpha's "series expansion" suspicious. The requirement for that to be valid are that the function in question be differentiable about the point in question (0, in this case), and this function is not defined at 0.

Also: it seems that there are two expressions being bantered about here. The original expression was

$\lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e + \frac{ex}2}{x^2}$

The other is
$\lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e - \frac{ex}2}{x^2}$

Which one is correct?
The correct expression is $\ \displaystyle \lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e + \dfrac{e \cdot x}2}{x^2}$ .

Although the function, $\left(1+x\right)^{(1/x)}$ is not defined at $x=0$, this is a removable discontinuity.

The following function is defined and differentiable at $x=0$, and has the Taylor expansion given by WolframAlpha for the function, $\left(1+x\right)^{(1/x)}$.

$\displaystyle \quad \quad u(x) = \begin{cases} \left(1+x\right)^{(1/x)} & \text{if } x \ne 0 \\ e & \text{if } x = 0 \end{cases}$

.

#### Physics lover

The correct expression is $\ \displaystyle \lim_{x \to 0} \dfrac{\left(1+x\right)^{(1/x)} - e + \dfrac{e \cdot x}2}{x^2}$ .

Although the function, $\left(1+x\right)^{(1/x)}$ is not defined at $x=0$, this is a removable discontinuity.

The following function is defined and differentiable at $x=0$, and has the Taylor expansion given by WolframAlpha for the function, $\left(1+x\right)^{(1/x)}$.

$\displaystyle \quad \quad u(x) = \begin{cases} \left(1+x\right)^{(1/x)} & \text{if } x \ne 0 \\ e & \text{if } x = 0 \end{cases}$

.
Do you have any idea to solve it without taylor expansion.

#### SammyS

Staff Emeritus
Homework Helper
Gold Member
Do you have any idea to solve it without Taylor expansion.
I don't have any good idea for that, but I'm thinking ...

As has been pointed out, the derivatives of the numerator get very complicated.

#### vela

Staff Emeritus
Homework Helper
Did you try L'Hopital's rule?

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