# Thread: Limit Marathon

1. Find $\lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$

2. Originally Posted by Krizalid
Find $\lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
$a^x=e^{\ln(a)x}=1+\ln(a)x+\frac{\ln^2(a)x^2}{2}+.. .$

$a^{-x}=1-\ln(a)x+\frac{\ln(a)^2x^2}{2}-...$

$a^{x}-a^{-x}=2\ln(a)x+...\sim{2\ln(a)x}$

and then we have

$log_a(a-x)=\frac{\ln(a-x)}{\ln(a)}=\frac{1}{\ln(a)}\bigg[\ln(a)+\ln\bigg(1-\frac{x}{a}\bigg)\bigg]=$ $\frac{1}{\ln(a)}\bigg[\ln(a)-\bigg(\frac{x}{a}\bigg)-\bigg(\frac{x}{a}\bigg)^2-...\bigg]=1-\frac{\bigg(\frac{x}{a}\bigg)}{\ln(a)}-\frac{\bigg(\frac{x}{a}\bigg)^2}{\ln(a)}$

So then

$1-x-\log_a(a-x)=1-x-1+\frac{\bigg(\frac{x}{a}\bigg)}{\ln(a)}+\frac{\bi gg(\frac{x}{a}\bigg)^2}{\ln(a)}+...$

So from that we can see that

$1-x-\log_a(a-x)\sim\frac{-(a\ln(a)-1)x}{a\ln(a)}$

$\therefore\lim_{x\to{0}}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)}\sim\lim_{x\to{0}}\frac{2\ln(a)x}{\frac{-(a\ln(a)-1)x}{a\ln(a)}}=\frac{-2a\ln^2(a)}{a\ln(a)-1}$

Thanks a lot Krizalid! I really enjoyed that one. It was messy, but still very fun.

3. Originally Posted by Krizalid
Find $\lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
Strangely L'Hospital gives a different answer from Mathstud's answer:

$\lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)} = \lim_{x\to0}\frac{a^x \ln a+ a^{-x}\ln a}{-1+\frac1{\ln a (a-x)}} = \frac{2a \ln ^2 a}{1 - a\ln a}$

I must have missed something

4. Originally Posted by Isomorphism
Strangely L'Hospital gives a different answer from Mathstud's answer:

$\lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)} = \lim_{x\to0}\frac{a^x \ln a+ a^{-x}\ln a}{-1+\frac1{\ln a (a-x)}} = \frac{2a \ln ^2 a}{1 - a\ln a}$

I must have missed something
No error! Look at my fourth step about half way in! I typoed a minus to a plus in the natural log, its my fault, I do these things on the computer not on pencil and paper, so when I make a typo it carries through the rest of the calculation

So yes you are right, my work will have a negative sign differenece

5. Originally Posted by Mathstud28
Here is another easy one

$\lim_{x\to{0}}x^{x^{x^{x^{x^{x^{x^x}}}}}}$
Hint: Think about pairs,

6. How 'bout a clean solution to my last problem?

7. Originally Posted by Krizalid
How 'bout a clean solution to my last problem?
You mean a non-power series solution? I acutally didnt look at this and see anything else but that, but if there is another sort of solution please say so and I will commence the search for it

8. Originally Posted by Krizalid
How 'bout a clean solution to my last problem?
How do you define a clean solution? Without LHospitals, power series?

Lets use only the definition of a derivative:

Then lets use the fact that $f(x) = a^x \Rightarrow \lim_{x \to 0} \frac{a^x - 1}{x} = f'(0) = \lim_{-x \to 0}\frac{a^{-x} - 1}{-x} = \lim_{x \to 0}\frac{a^{-x} - 1}{-x}$

Now $g(x) = \ln x \Rightarrow \lim_{x \to 0} \frac{\ln a - \ln (a-x)}{x} = g'(a) = \frac1{a}$

So $\lim_{x \to 0}\frac{a^x - a^{-x}}{1 - x - \log_a (a - x)} = \lim_{x \to 0}\dfrac{\frac{a^x - 1 -(a^{-x} - 1)}{x}}{\frac{\ln a - x\ln a - \ln (a - x)}{x\ln a}}=\lim_{x \to 0} \dfrac{f'(0) - (-f'(0))}{\frac{g'(a)}{\ln a} - 1}$

So $\lim_{x \to 0} \dfrac{f'(0) - (-f'(0))}{\frac{g'(a)}{\ln a} - 1} = \frac{2 \ln a}{\frac1{a\ln a} - 1} = \frac{2 a \ln^2 a}{1 - a\ln a}$

9. Popular Problem:

$\lim_{m \to \infty}\int_{0}^{\pi}\frac{\sin t}{1+\cos ^2 mt}\, dt$

10. Originally Posted by Isomorphism

How do you define a clean solution? Without LHospitals, power series?
... and no derivatives, only a solution which involves basic limits and some of algebra.

11. Originally Posted by Krizalid
... and no derivatives, only a solution which involves basic limits and some of algebra.
Isnt that unfair because every derivative "involves basic limits and some of algebra"?

I can understand what you are trying to say only if I see your solution perhaps

12. Originally Posted by Krizalid
Find $\lim_{x\to0}\frac{a^x-a^{-x}}{1-x-\log_a(a-x)},\,a>0,\,a\ne1.$
we only need two basic limits $\lim_{x\to0}\frac{a^x - 1}{x}=\ln a$ and $\lim_{x\to0}\frac{\log_a(1 - \frac{x}{a})}{x}=\frac{-1}{a\ln a}$, which are direct results of the definition

$\lim_{t\to{\pm \infty}}\left(1+\frac{1}{t}\right)^t=e$: for the first limit put $a^x - 1 =\frac{1}{t}$ and for the second limit put $\frac{-x}{a}=\frac{1}{t}.$ now we have:

$f(x)=\frac{a^x - a^{-x}}{1-x-\log_a(a-x)}=-a^{-x}(1+a^x) \times \frac{\frac{a^x -1}{x}}{1+\frac{\log_a(1-\frac{x}{a})}{x}}.$ thus: $\lim_{x\to0}f(x)=(-2) \times \frac{\ln a}{1 - \frac{1}{a\ln a}}=\frac{2a \ln^2 a}{1 - a\ln a}. \ \ \ \square$

13. Originally Posted by PaulRS
Let: $
a_1 \left( x \right) = x,a_2 \left( x \right) = x^x ,...,a_n \left( x \right) = \left. {x^{x^{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu
\raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1 mu}} ^x } } } \right\}{\text{ }}x{\text{ appears n times }}
$
be a sequence of functions

We'll show that: $
\mathop {\lim }\limits_{x \to 0^+} a_n \left( x \right) = \left\{ \begin{gathered}
0{\text{ if }}n \ne \dot 2 \hfill \\
1{\text{ if }}n = \dot 2 \hfill \\
\end{gathered} \right.
$

We have: $
\ln \left( {a_n } \right) = a_{n - 1} \cdot \ln \left( x \right)
$

The base cases $a_1(x)$ and $a_2(x)$ are quite easy, now consider $
n > 2
$

$
{\text{if }}n \ne \dot 2 \Rightarrow \mathop {\lim }\limits_{x \to 0^+} \ln \left( {a_n } \right) = \mathop {\lim }\limits_{x \to 0^+} \underbrace {a_{n - 1} }_{ \to 1} \cdot \ln \left( x \right) = - \infty
$

Thus we must have: $
a_n \to 0
$

$
{\text{if }}n = \dot 2 \Rightarrow \mathop {\lim }\limits_{x \to 0^+} \ln \left( {a_n } \right) = \mathop {\lim }\limits_{x \to 0^+} \underbrace {a_{n - 1} }_{ \to 0^+} \cdot \ln \left( x \right)
$

$
u = \ln \left( x \right) \Rightarrow a_{n - 1} = e^{u \cdot \overbrace {e^{u \cdot {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu
\raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1 mu}} ^{e^u } } }^{ \to C \in \mathbb{R}^ + }} \mathop \approx \limits_{u \to - \infty } e^{C \cdot u}
$

So that: $
\mathop {\lim }\limits_{x \to 0^+} \ln \left( {a_n } \right) = \mathop {\lim }\limits_{u \to - \infty } e^{C \cdot u} \cdot u = 0\therefore a_n \to 1
$

This completes the induction

We are asked $
\mathop {\lim }\limits_{x \to 0^+} a_8 \left( x \right) = 1
$
Good job Paul, here was my solution

Consider a function

$f(\xi)=\underbrace{\xi^{\xi^{\xi^{\cdots}}}}_{\tex t{An even number of times}}$

Now let $\psi=\xi^{\xi}$

So now we must find where $\psi$ goes as $\xi$ goes to zero

Which means we must find $\lim_{\xi\to{0}}\xi^{\xi}$

Now by letting $\xi=\frac{1}{\varphi}$

we get

$\lim_{\varphi\to\infty}\bigg(\frac{1}{\varphi}\big g)^{\frac{1}{\varphi}}$

Which by the realtion of the Ratio test and Root test is equivalent to

$\lim_{\varphi\to\infty}\frac{\frac{1}{\varphi+1}}{ \frac{1}{\varphi}}=lim_{\varphi\to\infty}\frac{\va rphi}{\varphi+1}=1$

$\therefore\text{ As }\xi\to{0}\Rightarrow\psi\to{1}$

Giving us

$\lim_{\psi\to{1}}\underbrace{\psi^{\psi^{\psi^{\cd ots}}}}_{\text{Half number of terms}}=1^{1^{1^{\cdots}}}=1$

$\therefore\lim_{\xi\to{0}}f(\xi)=1$

And now seeing that this limit falls under the reign of $f(\xi)$

The limit is one

14. Originally Posted by Isomorphism
Popular Problem:

$\lim_{m \to \infty}\int_{0}^{\pi}\frac{\sin t}{1+\cos ^2 mt}\, dt$
i think you're assuming that m is an integer. the idea of my solution is very simple but it requires

some writing ... so i wouldn't be surprised if some typos were found in it! ok, first see that:

$(1) \ \ \ \int_{(k-1)\pi}^{k\pi}\frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx = \int_0^{\pi} \frac{\sin(\frac{k\pi - x}{m})}{1+\cos^2x} \ dx.$ now, in your integral, let $mt = x.$ then:

$I_m=\int_0^{\pi} \frac{\sin t}{1+\cos^2 mt} \ dt=\frac{1}{m}\int_0^{m\pi} \frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx=\frac{1}{m} \sum_{k=1}^m \int_{(k-1)\pi}^{k\pi}\frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx.$ thus by $(1):$

$(2) \ \ \ I_m=\frac{1}{m}\sum_{k=1}^m \int_0^{\pi}\frac{\sin(\frac{k\pi - x}{m})}{1+\cos^2x} \ dx =$

$\frac{1}{m}\sum_{k=1}^m \sin(\frac{k\pi}{m})\int_0^{\pi} \frac{\cos(\frac{x}{m})}{1+\cos^2x} \ dx -\frac{1}{m}\sum_{k=1}^m\cos(\frac{k\pi}{m})\int_0^ {\pi}\frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx.$

now since $0 \leq \sin(\frac{x}{m}) \leq \frac{x}{m},$ and $0 \leq \sin^2(\frac{x}{2m}) \leq \frac{x^2}{4m^2}$, we'll have:

$(3) \ \ \ \lim_{m\to\infty} \int_0^{\pi} \frac{\sin(\frac{x}{m})}{1+\cos^2x} \ dx = 0,$ and:

$(4) \ \ \ \lim_{m\to\infty} \int_0^\pi \frac{\cos(\frac{x}{m})}{1+\cos^2x} \ dx=\lim_{m\to\infty} \int_0^{\pi} \frac{1-2\sin^2(\frac{x}{2m})}{1+\cos^2x} \ dx = \int_0^{\pi} \frac{1}{1+\cos^2x} \ dx=\frac{\pi}{\sqrt{2}}.$

it's also clear, by Riemann sum, that:

$(5) \ \ \ \lim_{m\to\infty}\frac{1}{m} \sum_{k=1}^m \sin(\frac{k\pi}{m}) = \frac{1}{\pi}\int_0^{\pi}\sin x \ dx = \frac{2}{\pi},$ and:

$(6) \ \ \ \lim_{m\to\infty} \frac{1}{m}\sum_{k=1}^m \cos(\frac{k\pi}{m}) = \frac{1}{\pi}\int_0^{\pi} \cos x \ dx = 0.$

thus from (2), (3), (4), (5), and (6) we have: $\lim_{m\to\infty} I_m = \sqrt{2}. \ \ \ \square$

15. Here is another one I came up with, find L.

$L=\lim_{x\to{0^{+}}}\bigg(a^{\frac{b}{x}+c}+f(x)+g (x)+...\bigg)^x$

Where $a^{bx}\succ{f\bigg(\frac{1}{x}\bigg)}\succ{g\bigg( \frac{1}{x}\bigg)}\succ\cdots$

Restrictions: You may not use asymptotic equivalences, rate of increase, or in any other way disregard terms.

EDIT: typo, also assuming there is no domain conflictions with other functions

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