## Limit approaching infinity with an unknown in the exponent

Limits, differentiation, related rates, integration, trig integrals, etc.
jimmy_boots
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### Limit approaching infinity with an unknown in the exponent

Find the limit, as x approaches infinity, of (1+(2/x))x

This is a problem from Temple University's 'Calculus on the Web' site, I am trying to get re-acquainted with first-semester Calculus as I will begin taking Calc II in a month from now. I can't think of how to bring that x down. Logs?

Thanks,

Jimmy

DAiv
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### Re: Limit approaching infinity with an unknown in the exponent

Find the limit, as x approaches infinity, of (1+(2/x))x

This is a problem from Temple University's 'Calculus on the Web' site, I am trying to get re-acquainted with first-semester Calculus as I will begin taking Calc II in a month from now. I can't think of how to bring that x down. Logs?
Do you need to bring it down?

If you plug infinity into x, starting from inside the parentheses and working outwards, what do you end up with?

DAiv

jimmy_boots
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### Re: Limit approaching infinity with an unknown in the exponent

2 divided by infinity goes to 0. 1 plus 0 is one. One raised to infinity is an indeterminate form, hence my wish to change the nature of the exponent x.

DAiv
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### Re: Limit approaching infinity with an unknown in the exponent

2 divided by infinity goes to 0. 1 plus 0 is one. One raised to infinity is an indeterminate form, hence my wish to change the nature of the exponent x.
$\begin{array} 1^0=1\\ 1^1=1\\ 1^2=1\\ 1^3=1\\ .\\ .\\ .\\ 1^\infty=?\\ \end{array}$

DAiv

jimmy_boots
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### Re: Limit approaching infinity with an unknown in the exponent

That's not helping me dave. It looks good intuitively but 1 is not a valid answer. I repeat, one raised to infinity is considered an indeterminate form. Maybe someone else would like to chime in?

DAiv
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### Re: Limit approaching infinity with an unknown in the exponent

That's not helping me dave. It looks good intuitively but 1 is not a valid answer. I repeat, one raised to infinity is considered an indeterminate form.
You're right, it's not a simple solution.

Q. Find $\lim_{x\to\infty}\left(1+\frac2x\right)^x$

There are at least two ways of solving this, a long, drawn out way, and a much simpler way.

The long, drawn out method

You need to use L'Hôpital's rule (twice) to convert the indeterminate forms into determinate ones. This involves converting the limit to a fractional form and deriving.

Basically, you need to:
1. Set the whole limit equal to a variable, say, y
2. Take the natural logarithm of both sides
3. Regroup, so the limit is out front
4. Rearrange the natural logarithm to bring down the power
5. Divide through to get a term containing 'x' on both top and bottom
(This gives a limit with the form $\frac00$ = indeterminate)
6. Use L'Hôpital's rule to circumvent this indeterminate form
(The limit will still have a term containing 'x' on both top and bottom, giving a limit with the form $\frac\infty\infty$ = indeterminate)
7. Use L'Hôpital's rule again to circumvent this last indeterminate form
8. Finally, solve for y

Here's the solution:
Spoiler:
Q. Find $\lim_{x\to\infty}\left(1+\frac2x\right)^x$

[1.] Set the whole limit equal to y,

$\text{Let }y = \lim_{\small x\to\infty} (1+\frac2x)^x$

[2.] Take the natural logarithm of both sides,

$\ln\, y = \ln\,\left[\lim_{\small x\to\infty} (1+\frac2x)^x\right]$

[3.] Since $\ln\,\left[\lim_{\small x\to\infty} (1+\frac2x)^x\right]$ is a continuous function, we can swap the positions of the limit and the natural logarithm,

$\ln\, y = \lim_{\small x\to\infty} \left[\ln\,(1+\frac2x)^x\right]$

[4.] Rearrange the natural logarithm to bring down the power,

$\ln\, y = \lim_{\small x\to\infty} \left[x \, \ln\,(1+\frac2x)\right]$

[5.] Divide through to get a term containing 'x' on both top and bottom,

$\ln\, y = \lim_{\small x\to\infty} \left[\frac{\ln\,(1+\frac2x)}{\frac1x}\right]$

This ultimately resolves to $\frac00 = indeterminate$, so...

[6.] Use L'Hôpital's rule to circumvent this indeterminate form,

a) Take the derivative of the numerator,

$\begin{eqnarray} &\frac{d}{dx}& \left[\ln\,(1+\frac2x)\right]\\ =&\frac{d}{dx}& \left[\ln\,(1+2x^{\small{-1}})\right]\\ =&&\frac{-2(x^{\small{-2}})}{(1+\frac2x)}\\ =&&\frac{-2}{x^2(1+\frac2x)}\\ =&&\frac{-2}{x(x+2)}\\ \end{eqnarray}$

b) Take the derivative of the denominator,

$\begin{eqnarray} &\frac{d}{dx}& \left[\frac1x\right]\\ =&\frac{d}{dx}& \left[x^{\small{-1}}\right]\\ =&&-1(x^{\small{-2}})\\ =&&-\frac1{x^2}\\ \end{eqnarray}$

c) Divide the derivative of the numerator by the derivative of the denominator,

$\begin{eqnarray} &\frac{\left[\frac{-2}{x(x+2)}\right]}{\left[-\frac1{x^2}\right]}\\ =&\left[\frac{-2}{x(x+2)}\right]\left[-\frac{x^2}1\right]\\ =&\frac{2x^2}{x(x+2)}\\ =&\frac{2x}{x+2}\\ \end{eqnarray}$

Substitute the result back into our limit,

$\begin{eqnarray} \ln\,y =&\lim_{x\to\infty}\left(\frac{2x}{x+2}\right)\\ \end{eqnarray}$

This ultimately resolves to $\frac\infty\infty = indeterminate$, so...

[7.] Use L'Hôpital's rule again to circumvent this last indeterminate form,

$\begin{eqnarray} \ln\,y =&\lim_{x\to\infty}\left[\frac21\right]\\ =&2\\ \end{eqnarray}$

[8.] Solve for y,

$\begin{eqnarray} \ln\, y = &2\\ \log_e{y} = &2\\ y =&e^2 \end{eqnarray}$

Therefore, $\lim_{x\to\infty}\left(1+\frac2x\right)^x = e^2$.

The simpler method

This uses the fact that $\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$ looks very similar to our given limit, $\lim_{x\to\infty}\left(1+\frac2x\right)^x$.

Here's the solution:
Spoiler:
Q. Find $\lim_{x\to\infty}\left(1+\frac2x\right)^x$

$\lim_{x\to\infty}\left(1+\frac2x\right)^x$

$\text{Let } x = 2u$

Substitute $2u$ for $x$,

$\begin{eqnarray}&\lim_{2u\to\infty}\left(1+\frac2{2u}\right)^{2u}\\ =&\lim_{u\to\infty}\left(1+\frac1u\right)^{u.2}\\ =&\left[\lim_{u\to\infty}\left(1+\frac1u\right)^u\right]^2 \end{eqnarray}$

But $\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$, so

$\begin{eqnarray} &\left[\lim_{u\to\infty}\left(1+\frac1u\right)^u\right]^2\\ =&[e]^2\\ =&e^2 \end{eqnarray}$

Therefore, $\lim_{x\to\infty}\left(1+\frac2x\right)^x = e^2$.
DAiv

jimmy_boots
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### Re: Limit approaching infinity with an unknown in the exponent

That's what I'm talking about! Thanks.

I follow the L'Hopital's Rule method of solving this problem, but where did you get the "$\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$" from?

This is a given identity thinger I should be familiar with? I haven't come across it, please let me know the deal.

Thanks again,

Jimmy

DAiv
Posts: 35
Joined: Tue Dec 16, 2008 7:47 pm
Contact:

### Re: Limit approaching infinity with an unknown in the exponent

I follow the L'Hopital's Rule method of solving this problem, but where did you get the "$\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$" from?

This is a given identity thinger I should be familiar with? I haven't come across it, please let me know the deal.
This is just one of many ways of representing $e$. It would be unreasonable to learn them all by heart, but a familiarisation with their general forms may alert you to the possibility of $e$ hiding in a particular problem. Learning a few of the simpler ones, like this, may prove useful (and save a lot of time).

In fact, by using the more general version of this representation of $e$, the second solution can be simplified further:

Q. Find $\lim_{x\to\infty}\left(1+\frac2x\right)^x$

Since $\lim_{x\to\infty}\left(1+\frac{y}x\right)^x = e^y$,

$\text{Let }y = 2$
Therefore, $\lim_{x\to\infty}\left(1+\frac{2}x\right)^x = e^2$.

Thinking about where this question originally came from, Temple University's 'Calculus on the Web' site, and the simplicity of the other 22 questions leading up to this one, I would imagine that the question may be testing for knowledge of this very representation of $e$.

DAiv