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?

Relevant 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.

 

[fresh_42]

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]

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]

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]

The answer is -11e/24.

Not according to WolframAlpha:
https://www.wolframalpha.com/input/?i=limit+(x+to+0)+((((1+x)^(1/x))-e-+(ex/2))/(x^2))

 

[Physics lover]

Not according to WolframAlpha:
https://www.wolframalpha.com/input/?i=limit+(x+to+0)+((((1+x)^(1/x))-e-+(ex/2))/(x^2))

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]

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]

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.

I am unable to converge that please help.

 

[statdad]

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]

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:
https://www.wolframalpha.com/input/?i=limit+(x+to+0)+((((1+x)^(1/x))-e+(ex/2))/(x^2))

 

[statdad]

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]

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.

 

[Physics lover]

See post #11.

It is not yet solved.Can you help fresh.

 

[fresh_42]

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.

 

[SammyS]

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]

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]

Did you try L'Hopital's rule?

 

[Physics lover]

Did you try L'Hopital's rule?

Yes but the terms are complicated.

 

[vela]

It's not too bad.

Let $f(x) = (1+x)^{1/x} = e^{g(x)}$ where $g(x) = \frac{\log (1+x)}{x}$. Then you have
\begin{align*}
f'(x) &= f(x)g'(x) \\
f''(x) &= f'(x)g'(x) + f(x) g''(x).
\end{align*} You already know the limit of $f$ as $x \to 0$, so to find the limit of $f'$, you just need to find the limit of $g'$ as $x \to 0$, which you can find using L'Hopital's rule. Once you have that, you can find the limit of $f''$ by finding the limit of $g''$, which, again, you can find using L'Hopital's rule.

 

[Physics lover]

It's not too bad.

Let $f(x) = (1+x)^{1/x} = e^{g(x)}$ where $g(x) = \frac{\log (1+x)}{x}$. Then you have
\begin{align*}
f'(x) &= f(x)g'(x) \\
f''(x) &= f'(x)g'(x) + f(x) g''(x).
\end{align*} You already know the limit of $f$ as $x \to 0$, so to find the limit of $f'$, you just need to find the limit of $g'$ as $x \to 0$, which you can find using L'Hopital's rule. Once you have that, you can find the limit of $f''$ by finding the limit of $g''$, which, again, you can find using L'Hopital's rule.

Sir i tried the same way you did but the answer is not coming.By the way i will try once more.

 

[Physics lover]

It's not too bad.

Let $f(x) = (1+x)^{1/x} = e^{g(x)}$ where $g(x) = \frac{\log (1+x)}{x}$. Then you have
\begin{align*}
f'(x) &= f(x)g'(x) \\
f''(x) &= f'(x)g'(x) + f(x) g''(x).
\end{align*} You already know the limit of $f$ as $x \to 0$, so to find the limit of $f'$, you just need to find the limit of $g'$ as $x \to 0$, which you can find using L'Hopital's rule. Once you have that, you can find the limit of $f''$ by finding the limit of $g''$, which, again, you can find using L'Hopital's rule.

Thanks for the help sir.Finally I got the answer but it took too much time to find the limit g"(x).For that i used L'Hospital 3 times.