Results 1 to 7 of 7

Math Help - Some hard limits I can't solve

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    7

    Some hard limits I can't solve

    Hello, I've been solving limits because I have a test in a few days, but a few of them i really cant find... (I study for engineer in Belgium). I expect them to be at least this hard, because they want about 2/3 to fail the 1st year!

    Limit x to infiny : ((e^2x) -5x)^(1/x)
    Limit x to infinty: x(ln(x+1)-ln(x))

    Also, I know limit x to 0: sinx/x is 1, but limit x to zero abs(sinx)/x ?

    Thank you in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kev91 View Post
    Hello, I've been solving limits because I have a test in a few days, but a few of them i really cant find... (I study for engineer in Belgium). I expect them to be at least this hard, because they want about 2/3 to fail the 1st year!

    Limit x to infiny : ((e^2x) -5x)^(1/x)
    Limit x to infinty: x(ln(x+1)-ln(x))

    Also, I know limit x to 0: sinx/x is 1, but limit x to zero abs(sinx)/x ?

    Thank you in advance.
    The technique for the first is exemplified in this thread: http://www.mathhelpforum.com/math-he...ate-limit.html

    For the second, note that x ( \ln (x + 1) - \ln (x)) = \frac{\ln (x + 1) - \ln (x)}{\frac{1}{x}} and now use l'Hopital's Rule.

    For the third, the left hand limits and right hand limits are different (-1 and 1) therefore the limit does not exist.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,711
    Thanks
    630
    Hello, kev91!

    The second can be solved with this defintion: . \lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^x \;=\;e


    2)\;\;\lim_{x\to\infty}x\bigg[\ln(x+1)-\ln(x)\bigg]

    We have: . \lim_{x\to\infty} x\ln\left(\frac{x+1}{x}\right) \;=\;\lim_{x\to\infty}x\left(1 + \tfrac{1}{x}\right) \;=\;\lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^  x \;=\;e

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Soroban View Post
    Hello, kev91!

    The second can be solved with this defintion: . \lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^x \;=\;e


    We have: . \lim_{x\to\infty} x {\color{red}\ln} \left(\frac{x+1}{x}\right) \;=\;\lim_{x\to\infty}x\left(1 + \tfrac{1}{x}\right) \;=\;\lim_{x\to\infty}\left(1+\tfrac{1}{x}\right)^  x \;=\;e
    Clearly you're not a lumber jack, Soroban. Because you dropped a (red) log ....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2009
    Posts
    7
    Mr fantastic, the 1st limit is to infinity so, I cant use that method you showed me.

    Also, I'm prohibitit to use Lhopital for the 2nd limit, but I found it the way Soroban showed me, but a bit differently because he dropped a ln for no reason , the solution is 1. (ln(e))
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by kev91 View Post
    Mr fantastic, the 1st limit is to infinity so, I cant use that method you showed me. Mr F says: *Ahem* Yes you can. The indeterminant form is {\color{red}\frac{\infty}{\infty}}. See - Wolfram|Alpha and be sure to click on Show steps.

    Also, I'm prohibitit to use Lhopital for the 2nd limit, but I found it the way Soroban showed me, but a bit differently because he dropped a ln for no reason , the solution is 1. (ln(e))
    Normally I'm not lazy enough to suggest l'Hopital's Rule. But since it is a legitimate method .... (next time, try to mention techniques that you're NOT allowed to use). I suppose you're going to say that you're not allowed to l'Hopital's Rule for the first one either .... (In which case you can note that \ln (e^{2x} - 5x) ~ 2x ....)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2009
    Posts
    7
    Ok thank you, you have been really helpfull! and yes, I was allowed to use l'hopital in the 1st one .

    For this one I have to use Lhopital, but its kind of stupid because I solved it without l'hopital.
    lim x to 0 : x^sinx = e ^ lim x to 0: sin x *ln x

    sin 0 = 0

    and when ln x approaches x=0 is also 0, so the solution is e^0 and that's 1, but it says i HAVE to use l'hopital rule.

    Also can you use lhopital when you have 0/ infinity, or does it have to be 0/0 or infinity/infinity.

    Sorry for my stupid questions, but they just explained the basics during the class =(, and the test are always way harder!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coin tossing and Cards Experiment Which I Find Hard To Solve
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 13th 2010, 04:34 PM
  2. A integral problem hard to solve
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 30th 2009, 10:32 AM
  3. Replies: 3
    Last Post: May 31st 2008, 05:45 AM
  4. Replies: 0
    Last Post: February 11th 2008, 03:22 AM
  5. any one help.....so hard to solve...!!!
    Posted in the Number Theory Forum
    Replies: 10
    Last Post: July 21st 2006, 07:19 AM

Search Tags


/mathhelpforum @mathhelpforum