Exploring Alternative Solutions to a Limit Equation

[Saitama]

Homework Statement

$$\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$
(its the whole term to the power 1/x, can somebody tell me the correct latex code?)

Homework Equations

The Attempt at a Solution

I have actually solved the problem but i am trying to find an alternative to do it.
My attempt:
(I am dropping the limit word just to make it easier for me to write)
let
$$y=\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$
$$lny=\frac{1}{x}[\frac{ln(1+x)-x}{x}]$$
Using the series of ln(1+x), taking out the common factor x^2 and applying the limit, i get
$$lny=\frac{-1}{2}$$
$$y=e^{\frac{-1}{2}}$$

Is it possible that i can solve it without using the series expansion of ln(1+x)? Also, i am not allowed to use the L'Hôpital's rule.

Thanks!

 

[Mark44]

Homework Statement

$$\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$
(its the whole term to the power 1/x, can somebody tell me the correct latex code?)

Instead of using \stackrel, just use \lim_{x → 0}, like this:
$$ \lim_{x \to 0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$

Homework Equations

The Attempt at a Solution

I have actually solved the problem but i am trying to find an alternative to do it.
My attempt:
(I am dropping the limit word just to make it easier for me to write)
let
$$y=\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$

Let y = the expression inside the limit. Don't include the limit at this step.

$$lny=\frac{1}{x}[\frac{ln(1+x)-x}{x}]$$

The above might be wrong. You have skipped quite a few steps here, so I'm not following what you did.

What happened to the inner 1/x exponent? What happened to the e in the denominator?

Using the series of ln(1+x), taking out the common factor x^2 and applying the limit, i get
$$lny=\frac{-1}{2}$$
$$y=e^{\frac{-1}{2}}$$

Is it possible that i can solve it without using the series expansion of ln(1+x)? Also, i am not allowed to use the L'Hôpital's rule.

Thanks!

This really should go in the Calc & Beyond section - I'm moving it there.

 

[Infinitum]

$$lny=\frac{1}{x}[\frac{ln(1+x)-x}{x}]$$

The above might be wrong. You have skipped quite a few steps here, so I'm not following what you did.

What happened to the inner 1/x exponent? What happened to the e in the denominator?

Quite sure this is correct, as I get the same expression after simplification. Using logarithm properties, 1/e becomes -1 and the expression boils down to this.

Is it possible that i can solve it without using the series expansion of ln(1+x)?

Probably. What I did is,

$$\lim_{x\to0} (1+f(x))^{g(x)} = e^{\lim_{x\to0} f(x)g(x)}$$

Which is applicable iff g(x) tends to infinity, which it does.

From this I get,

$$e^{\lim_{x\to0} \frac{(1+x)^{1/x}-e}{ex}}$$

(Edit : Probably need to zoom to see this one. Here's the exponent part)

$${\lim_{x\to0} \frac{(1+x)^{1/x}-e}{ex}}$$
Finding the limit of this function will give you the answer. You can probably do it by general binomial series, but I'm trying to find a way to get this limit without using L'Hospital Rule or those series.

 

[cjc0117]

The above might be wrong. You have skipped quite a few steps here, so I'm not following what you did.

What happened to the inner 1/x exponent? What happened to the e in the denominator?

OP is right:

$y=lim_{x\rightarrow 0}(\frac{(1+x)^{1/x}}{e})^{1/x}$

$ln(y)=ln(lim_{x\rightarrow 0}(\frac{(1+x)^{1/x}}{e})^{1/x})$

$ln(y)=lim_{x\rightarrow 0}ln((\frac{(1+x)^{1/x}}{e})^{1/x})$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}ln(\frac{(1+x)^{1/x}}{e})$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(ln((1+x)^{1/x})-ln(e))$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(\frac{1}{x}ln(1+x)-1)$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(\frac{ln(1+x)-x}{x})$

 

[spamiam]

Homework Statement

$$\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$
(its the whole term to the power 1/x, can somebody tell me the correct latex code?)

Use \left and \right before your parentheses and brackets to have them automatically resize. You can click quote to check out the example below:
$$
\lim_{x \to 0} \left( \frac{(1+x)^{1/x}}{e} \right)^{1/x}
$$

The Attempt at a Solution

I have actually solved the problem but i am trying to find an alternative to do it.
My attempt:
(I am dropping the limit word just to make it easier for me to write)
let
$$y=\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$
$$lny=\frac{1}{x}[\frac{ln(1+x)-x}{x}]$$
Using the series of ln(1+x), taking out the common factor x^2 and applying the limit, i get
$$lny=\frac{-1}{2}$$
$$y=e^{\frac{-1}{2}}$$

Is it possible that i can solve it without using the series expansion of ln(1+x)? Also, i am not allowed to use the L'Hôpital's rule.

Thanks!

Your solution is correct (I used L'Hopital's Rule), but I don't see a way to do it without using L'Hopital's rule or series. You could also use the generalized binomial series I suppose, but that's no simpler than your series for log.

 

[Saitama]

Instead of using \stackrel, just use \lim_{x → 0}, like this:
$$ \lim_{x \to 0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$

Thanks for the code Mark!

The above might be wrong. You have skipped quite a few steps here, so I'm not following what you did.

What happened to the inner 1/x exponent? What happened to the e in the denominator?

Yes, i did skip a few steps as they involved only simple properties that are mentioned by cjc0117 and Infinitum.

This really should go in the Calc & Beyond section - I'm moving it there.

Why this should go in Calc and Beyond? I haven't used any calculus in this question. (Now i don't want to get the same infraction again)

Probably. What I did is,

$$\lim_{x\to0} (1+f(x))^{g(x)} = e^{\lim_{x\to0} f(x)g(x)}$$

Which is applicable if g(x) tends to infinity, which it does.

From this I get,

$$e^{\lim_{x\to0} \frac{(1+x)^{1/x}-e}{ex}}$$

(Edit : Probably need to zoom to see this one. Here's the exponent part)

$${\lim_{x\to0} \frac{(1+x)^{1/x}-e}{ex}}$$
Finding the limit of this function will give you the answer. You can probably do it by general binomial series, but I'm trying to find a way to get this limit without using L'Hospital Rule or those series.

Thanks for trying out the problem. I am asking for an alternative method as i tend to forget the series expansion and binomial which serves me no good in my exams.

OP is right:

$y=lim_{x\rightarrow 0}(\frac{(1+x)^{1/x}}{e})^{1/x}$

$ln(y)=ln(lim_{x\rightarrow 0}(\frac{(1+x)^{1/x}}{e})^{1/x})$

$ln(y)=lim_{x\rightarrow 0}ln((\frac{(1+x)^{1/x}}{e})^{1/x})$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}ln(\frac{(1+x)^{1/x}}{e})$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(ln((1+x)^{1/x})-ln(e))$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(\frac{1}{x}ln(1+x)-1)$

$ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(\frac{ln(1+x)-x}{x})$

Exactly what i did.

Use \left and \right before your parentheses and brackets to have them automatically resize. You can click quote to check out the example below:
$$
\lim_{x \to 0} \left( \frac{(1+x)^{1/x}}{e} \right)^{1/x}
$$

Thanks spamiam, i was in need of this.

 

[Mark44]

Why this should go in Calc and Beyond? I haven't used any calculus in this question. (Now i don't want to get the same infraction again)

Limits are more associated with calculus than with the precalc courses. Also, this type of problem lends itself to L'Hopitals's Rule, which requires the use of derivatives. In addition, in your approach, you used a series representation of ln(1 + x), which is another calculus concept. For all of these reasons, I moved the thread. I figured it was just an honest mistake on your part -- no infraction.

 

[Saitama]

Limits are more associated with calculus than with the precalc courses. Also, this type of problem lends itself to L'Hopitals's Rule, which requires the use of derivatives. In addition, in your approach, you used a series representation of ln(1 + x), which is another calculus concept. For all of these reasons, I moved the thread. I figured it was just an honest mistake on your part -- no infraction.

Thanks once again for the clarification, i will take care of that from now.

 

[tt2348]

Try using x=1/n , and taking lim n->Infinity