A limit problem without the use of a Taylor series expansion

In summary: WolframAlpha says differently:...In summary, the conversation revolves around attempting to solve a limit involving a complex function using various methods, including L'Hôpital's rule and the Taylor series. The discussion also addresses the issue of using LaTeX to make the function more easily readable and the difficulty of finding a solution without using Taylor series. Ultimately, it is concluded that the limit does not exist and a plot of the function shows a vertical asymptote as x approaches 0 from the right.
  • #1
Physics lover
249
25
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.
 
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  • #2
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?
 
  • #3
fresh_42 said:
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.
 
  • #4
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}##
 
  • #5
fresh_42 said:
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.
 
  • #6
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.
 
  • #7
I tried L'Hospital but it is becoming complicated.
 
  • #8
fresh_42 said:
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.
 
  • #11
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.
 
  • #12
fresh_42 said:
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.
 
  • #13
Physics lover said:
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.
 
  • #14
fresh_42 said:
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.
 
  • #15
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.
 
  • #16
statdad said:
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))
 
  • #17
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?
 
  • #18
statdad said:
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.
 
  • #19
fresh_42 said:
See post #11.
It is not yet solved.Can you help fresh.
 
  • #20
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:
  • #21
statdad said:
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} ##

.
 
  • #22
SammyS said:
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.
 
  • #23
Physics lover said:
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.
 
  • #24
Did you try L'Hopital's rule?
 
  • #25
vela said:
Did you try L'Hopital's rule?
Yes but the terms are complicated.
 
  • #26
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.
 
  • #27
vela said:
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.
 
  • #28
vela said:
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.
 

What is a limit problem without the use of a Taylor series expansion?

A limit problem without the use of a Taylor series expansion is a mathematical problem that involves finding the value that a function approaches as its input approaches a specific value. This can be solved without using a Taylor series expansion, which is a method of approximating a function using its derivatives.

Why is it important to solve limit problems without using a Taylor series expansion?

Solving limit problems without using a Taylor series expansion can be important because it allows us to find the exact value of a limit, rather than just an approximation. This can be especially useful in certain applications, such as in physics or engineering, where precise values are needed.

What are some alternative methods for solving limit problems without using a Taylor series expansion?

Some alternative methods for solving limit problems without using a Taylor series expansion include direct substitution, factoring, and using trigonometric identities. These methods can be used depending on the specific problem and the techniques that the individual is comfortable with.

Can limit problems without using a Taylor series expansion be solved for all types of functions?

Yes, limit problems without using a Taylor series expansion can be solved for all types of functions, including polynomial, exponential, logarithmic, and trigonometric functions. However, the difficulty of solving the problem may vary depending on the type of function and the specific limit being evaluated.

Are there any limitations to solving limit problems without using a Taylor series expansion?

One limitation of solving limit problems without using a Taylor series expansion is that it may not always be the most efficient or straightforward method. In some cases, using a Taylor series expansion may be quicker and easier to solve the problem. Additionally, some limit problems may require more advanced techniques that cannot be solved without using a Taylor series expansion.

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