Results 1 to 5 of 5

Math Help - Limit without using L'Hospital Rule

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    19

    Limit without using L'Hospital Rule

    I've already solved this porblem using L'Hospital rule, and now I'm required to solve it without using this rule, rather using the given condition.

    Find:

    Lim (sinx)^tanx
    x-->pi/2



    Given that: Lim (1+1/x)^x= e
    x-->infinity

    The answer is, as you might have guessed, 1. But, can anyone do it absolutely without using L'Hopital rule? I tried to do, but it was only elegant solution, so that's worthless.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    a) considering the identity...

    (\sin x)^{\tan x} = e^{\tan x\cdot \ln \sin x} (1)

    ... and the fact that is...

    \ln \sin x = \frac{1}{2}\cdot \ln (1- \cos^{2} x) = - \frac{1}{2}\cdot (\cos^{2} x + \frac{\cos ^{4} x}{2} + \dots) (2)

    ... so that is...

    \lim_{x \rightarrow \frac {\pi}{2}} \frac{\sin x\cdot \ln \sin x}{\cos x} = 0 (3)

    ... we conclude that is...

    \lim_{x \rightarrow \frac {\pi}{2}} (\sin x)^{\tan x} = 1 (4)

    b) considering that for x>1 is...

    \ln \{(1+\frac{1}{x})^{x}\} = x \cdot \ln (1+\frac{1}{x}) = x \cdot (\frac{1}{x} - \frac{1}{2x^{2}} + \frac{1}{3x^{3}} - \dots) (5)

    ... we conclude that is...

    \lim_{ x \rightarrow \infty} (1+\frac{1}{x})^{x} = e (6)

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 9th 2009 at 06:41 AM. Reason: Correction of a basic mistake... very sorry!...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    19
    Hey, your no. (2) is completely wrong. You get it in the form (0/0) not negative infinity. So the solution is just elegant. I just need to find the limit given that condition (meaning using that condition, instead of L'Hospital rule). So... if anyone can please help me.

    Quote Originally Posted by chisigma View Post
    a) considering the identity...

    (\sin x)^{\tan x} = e^{\tan x\cdot \ln \sin x} (1)

    ... and the fact that is...

    \lim_{x \rightarrow \frac {\pi}{2}} \frac{\sin x\cdot \ln \sin x}{\cos x} = - \infty (2)

    ... we conclude that is...

    \lim_{x \rightarrow \frac {\pi}{2}} (\sin x)^{\tan x} =0 (3)

    b) considering that for x>1 is...

    \ln \{(1+\frac{1}{x})^{x}\} = x \cdot \ln (1+\frac{1}{x}) = x \cdot (\frac{1}{x} - \frac{1}{2x^{2}} + \frac{1}{3x^{3}} - \dots) (4)

    ... we conclude that is...

    \lim_{ x \rightarrow \infty} (1+\frac{1}{x})^{x} = e (5)

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by concern View Post
    I've already solved this porblem using L'Hospital rule, and now I'm required to solve it without using this rule, rather using the given condition.

    Find:

    Lim (sinx)^tanx
    x-->pi/2



    Given that: Lim (1+1/x)^x= e
    x-->infinity

    The answer is, as you might have guessed, 1. But, can anyone do it absolutely without using L'Hopital rule? I tried to do, but it was only elegant solution, so that's worthless.
    Problem: Compute L=\lim_{x\to\frac{\pi}{2}}\sin(x)^{\tan(x)} without use of L'hopital's rule

    Solution (1): Let x=\frac{\pi}{2}-z so L=\lim_{z\to0}\sin\left(\frac{\pi}{2}-z\right)^{\tan\left(\frac{\pi}{2}-z\right)}=\lim_{z\to0}\cos(z)^{\cot(z)}. Finally we can see that L=\lim_{z\to0}\cos(z)^{\cot(z)}\implies\ln\left(L\  right)=\lim_{z\to0}\cot(z)\ln\left(\cos(z)\right). Now in the neighborhood of zero \cos(z)\stackrel{0}\sim 1-\frac{z^2}{2} so that \ln\left(L\right)=\lim_{z\to0}\cot(z)\ln\left(1-\frac{z^2}{2}\right) and lastly noting that \ln\left(1-\frac{z^2}{2}\right)\stackrel{0}\sim \frac{z^2}{2} we may conclude that \ln\left(L\right)=\lim_{z\to0}\frac{z^2}{2}\cot(z)  =\frac{1}{2}\lim_{z\to0}\frac{z^2}{\tan(z)}. And once again appealing to asymptotic equivalence \tan(z)\stackrel{0}\sim z so \ln\left(L\right)=\frac{1}{2}\lim_{z\to0}\frac{z^2  }{z}=0\implies L=e^0=1.


    Solution (2): Let \tan(x)=z so that L=\lim_{z\to\infty}\sin\left(\text{arctan}(z)\righ  t)^z=\lim_{z\to\infty}\left(\frac{z}{\sqrt{z^2+1}}  \right)^z. Letting z=\frac{1}{\ell} turns this limit into L=\lim_{\ell\to0}\left(\frac{\frac{1}{\ell}}{\sqrt  {\frac{1}{\ell^2}+1}}\right)^{\frac{1}{\ell}}=\lim  _{\ell\to0}\left(\frac{1}{\sqrt{\ell^2+1}}\right)^  \frac{1}{\ell}\implies\ln\left(L\right)=\frac{-1}{2}\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right)  }{\ell}. But \lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right)}{\el  l}=\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right)-\ln\left(1+0^2\right)}{\ell-0}=\bigg[ \ln\left(1+\ell^2\right)\bigg]'\bigg|_{\ell=0}=\bigg[\frac{2\ell}{\ell^2+1}\bigg]\bigg|_{\ell=0}=0 (why?). So that \ln\left(L\right)=\frac{-1}{2}\lim_{\ell\to0}\frac{\ln\left(1+\ell^2\right)  }{\ell}=\frac{-1}{2}\cdot0=0\implies L=e^0=1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    19
    Thanks Drexel, this works perfectly fine.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding Limit With L'Hospital's Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 30th 2011, 02:30 PM
  2. Applying l'Hospital's rule to find a limit
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: October 17th 2011, 06:23 AM
  3. Finding this limit without l'Hospital's rule
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 24th 2009, 04:00 PM
  4. Replies: 4
    Last Post: September 21st 2009, 01:28 PM
  5. Limit without L'Hospital's Rule.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 11th 2009, 01:02 PM

Search Tags


/mathhelpforum @mathhelpforum