1. Originally Posted by Krizalid
Try this one: $\lim_{\alpha\to\infty}\frac1{\alpha^2}\int_0^\alph a\ln(1+e^{x})\,dx.$

$\lim_{\alpha\to\infty}\frac{\int_0^{\alpha}\ln(1+e ^x)}{\alpha^2}=\lim_{\alpha\to\infty}\frac{\ln(1+e ^{\alpha})}{2\alpha}$

now $\ln(1+e^{\alpha})\sim\ln(e^{\alpha})=\alpha$

thus the limit is $\lim_{\alpha\to\infty}\frac{1}{\alpha^2}\int_0^{\a lpha}\ln(1+e^x)dx=\frac{1}{2}$

by L'hopitals

2. That's the correct answer, but find another solution without that rule.

3. Originally Posted by Krizalid
That's the correct answer, but find another solution without that rule.
Am I on the right track (whited out)

convert the integral to a Riemann sum?

4. $\alpha\cdot{\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}=\ int_0^{\alpha}\ln(1+e^x)dx$

Thus: $\frac{\int_0^{\alpha}\ln(1+e^x)dx}{\alpha^2}=\frac {\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}{\alpha}$

So our limit is: $\lim_{\alpha\rightarrow{+\infty}}\frac{\int_0^1\ln (1+e^{\alpha\cdot{x}})dx}{\alpha}=$ $\int_0^1\lim_{\alpha\rightarrow{+\infty}}\frac{\ln (1+e^{\alpha\cdot{x}})}{\alpha}dx=\int_0^1xdx$ $=\frac{1}{2}$

5. Originally Posted by PaulRS
$\alpha\cdot{\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}=\ int_0^{\alpha}\ln(1+e^x)dx$

Thus: $\frac{\int_0^{\alpha}\ln(1+e^x)dx}{\alpha^2}=\frac {\int_0^1\ln(1+e^{\alpha\cdot{x}})dx}{\alpha}$

So our limit is: $\lim_{\alpha\rightarrow{+\infty}}\frac{\int_0^1\ln (1+e^{\alpha\cdot{x}})dx}{\alpha}=$ $\int_0^1\lim_{\alpha\rightarrow{+\infty}}\frac{\ln (1+e^{\alpha\cdot{x}})}{\alpha}dx=\int_0^1xdx$ $=\frac{1}{2}$
Is there some theorem that says your first change with the integral works, or did you just find it by inpsection?

6. Okay, here's another solution:

$\int_{0}^{\alpha }{\ln \left( 1+e^{x} \right)\,dx}=\frac{1}{2}\alpha ^{2}+\int_{0}^{\alpha }{\ln \left( 1+e^{-x} \right)\,dx}.$

Since $\ln \left( 1+e^{-x} \right)$ is comparable to $e^{-x}$ for large $x,$ the second integral is bounded for all $\alpha,$ then as $\alpha\to\infty$ we have $\int_{0}^{\alpha }{\ln \left( 1+e^{x} \right)\,dx}=\frac{1}{2}\alpha ^{2}+\mathcal O(1)\implies \frac{1}{\alpha ^{2}}\int_{0}^{\alpha }{\ln \left( 1+e^{x} \right)\,dx}=\frac{1}{2}+\mathcal O\left( \alpha^{-2} \right),$ and the limit is $\frac12.$

Next problem: find $\lim_{n\to\infty}\int_0^\pi\bigg|\bigg(x+\frac\pi n\bigg)\sin nx\bigg|\,dx,\,n\in\mathbb N.$

7. Originally Posted by Mathstud28
Try this one without L'hopitals

$\lim_{\xi\to{0}}\frac{\ln(\sin(a\xi))}{\ln(\sin(b\ xi))}$
a little trick to remember:

$\lim_{\xi\to0} \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=1+\lim_{\x i\to0} \frac{\ln(\sin(a\xi))-\ln(\sin(b\xi))}{\ln(\sin(b\xi))}=1+\lim_{\xi\to0} \frac{\ln \left(\frac{\sin(a\xi)}{\sin(b\xi)}\right)}{\ln(\s in(b\xi))}$

$=1+\frac{\ln(\frac{a}{b})}{-\infty}=1. \ \ \ \square$
.................................................. ..................................................

note that in order to make sure everything is defined, we must either have

$a>0, \ b>0,$ and $\xi \rightarrow 0+$, or $a<0, \ b < 0,$ and $\xi \rightarrow 0-.$

8. Originally Posted by NonCommAlg
a little trick to remember:

$\lim_{\xi\to0} \frac{\ln(\sin(a\xi))}{\ln(\sin(b\xi))}=1+\lim_{\x i\to0} \frac{\ln(\sin(a\xi))-\ln(\sin(b\xi))}{\ln(\sin(b\xi))}=1+\lim_{\xi\to0} \frac{\ln \left(\frac{\sin(a\xi)}{\sin(b\xi)}\right)}{\ln(\s in(b\xi))}$

$=1+\frac{\ln(\frac{a}{b})}{-\infty}=1. \ \ \ \square$
.................................................. ..................................................

note that in order to make sure everything is defined, we must either have

$a>0, \ b>0,$ and $\xi \rightarrow 0+$, or $a<0, \ b < 0,$ and $\xi \rightarrow 0-.$
So your first move was to

$\text{add }1+-1=1+-\frac{\ln(\sin(b\xi)}{\ln(\sin(b\xi))}$

9. right Mathstud28, just add 1 and subtract 1. it's as simple as that!

Originally Posted by Krizalid
Next problem: find $\lim_{n\to\infty}\int_0^\pi\bigg|\bigg(x+\frac\pi n\bigg)\sin nx\bigg|\,dx,\,n\in\mathbb N.$
first of all the absolute value sign is needed for $\sin nx$ only.

clearly $\int_{k\pi}^{(k+1)\pi}y|\sin y| \ dt = \int_0^{\pi}(y+k\pi)\sin y \ dy.$ call this (1). now, in your integral, let $nx+\pi=y.$ then:

$I_n=\int_0^{\pi}\left(x + \frac{\pi}{n}\right)|\sin nx| \ dx=\frac{1}{n^2}\int_{\pi}^{(n+1)\pi}y|\sin y| \ dy=\frac{1}{n^2}\sum_{k=1}^n \int_{k\pi}^{(k+1)\pi}y|\sin y|\ dy.$ therefore by (1):

$I_n=\frac{1}{n^2}\sum_{k=1}^n \int_0^{\pi}(y+k\pi)\sin y \ dy=\frac{1}{n}\int_0^{\pi}y\sin y \ dy \ + \ \frac{n(n+1)\pi}{2n^2}\int_0^{\pi}\sin y \ dy.$ therefore:

$\lim_{n\to\infty}I_n=0 \ + \ \lim_{n\to\infty} \frac{n(n+1)\pi}{n^2}=\pi. \ \ \ \square$

10. That's correct.

11. Originally Posted by NonCommAlg
right Mathstud28, just add 1 and subtract 1. it's as simple as that!

first of all the absolute value sign is needed for $\sin nx$ only.

clearly $\int_{k\pi}^{(k+1)\pi}y|\sin y| \ dt = \int_0^{\pi}(y+k\pi)\sin y \ dy.$ call this (1). now, in your integral, let $nx+\pi=y.$ then:

$I_n=\int_0^{\pi}\left(x + \frac{\pi}{n}\right)|\sin nx| \ dx=\frac{1}{n^2}\int_{\pi}^{(n+1)\pi}y|\sin y| \ dy=\frac{1}{n^2}\sum_{k=1}^n \int_{k\pi}^{(k+1)\pi}y|\sin y|\ dy.$ therefore by (1):

$I_n=\frac{1}{n^2}\sum_{k=1}^n \int_0^{\pi}(y+k\pi)\sin y \ dy=\frac{1}{n}\int_0^{\pi}y\sin y \ dy \ + \ \frac{n(n+1)\pi}{2n^2}\int_0^{\pi}\sin y \ dy.$ therefore:

$\lim_{n\to\infty}I_n=0 \ + \ \lim_{n\to\infty} \frac{n(n+1)\pi}{n^2}=\pi. \ \ \ \square$
Have you seen this problem before..or do you just have a knack for seeing embarrassingly obvious little tricks?

Why dont you submit a limit NonCommonAlg? That would be great, for I am sure it would be fascinating...but noting your other posts..it would preferably NOT include any of the following:
Limits involving sums
Limits involving sums esecially those excluding odd primes
Limits involving integrals
Limits involving integrals that require both countour integration and a parameterazation to solve
Limits involving infinite products
Limits involving anything to do with Wallis's Product
I am sure there are quite a few more

12. Originally Posted by Mathstud28
Have you seen this problem before..or do you just have a knack for seeing embarrassingly obvious little tricks?
no i haven't, which means the "or" part of your question is probably true!

13. Originally Posted by NonCommAlg
no i haven't, which means the "or" part of your question is probably true!
So are you going to try to oblige that request...despite the inummerable restrictions I am sure will vex you?

14. Originally Posted by Mathstud28
So are you going to try to oblige that request...despite the inummerable restrictions I am sure will vex you?
i might! right now, i don't have any that survives your tough conditions! haha

15. Originally Posted by NonCommAlg
i might! right now, i don't have any that survives your tough conditions! haha
Haha, if you have any good limits, I would love to see them. So don't hesitate if you find any!

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