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Homework Help: How to solve these limits ?

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the limits of the following

    2. Relevant equations
    We can't use L'Hôpital's rule as it's not in our syllabus.We have to use only

    3. The attempt at a solution
    I attempted a lot but can't find a way to solve these three (out of about 80 limits)
    pls show me the steps to get these without L'Hôpital's as Wolfram Alpha give steps in that law...
  2. jcsd
  3. Jun 2, 2012 #2
    Mind showing us your attempts first?
    Nobody will help you if you don't show your efforts.

    Just to give you a start for the first question, can you make it of the form 1?
  4. Jun 2, 2012 #3
    The first one is [itex]\infty \cdot 0[/itex], and can be converted to an indeterminate form [itex]\frac{0}{0}[/itex] by representing [itex]\tan x = \frac{1}{\cot x}[/itex]. Then, apply L'Hospital's Rule to:
    \lim_{x \rightarrow \pi/2}{\frac{\ln (\sin x)}{\cot x}}
    Remember to use chain rule when finding the derivatives.

    EDIT: Ah, you can't apply LR. Take [itex]t \equiv \ln(\sin x), x \rightarrow \frac{\pi}{2} \Rightarrow t \rightarrow ?[/itex]. Then [itex]\sin x = e^{t}[/itex]. Use trigonometry to express [itex]\cot x[/itex].
  5. Jun 2, 2012 #4
    No idea about 1 form..

    log(sinx)=log10sinx or log(sinx)=logesinx ?
  6. Jun 2, 2012 #5
    Hey, it's your problem. You tell me. The solution changes very little regardless of the base of the lograrithm.
  7. Jun 2, 2012 #6
    It was given in that way...don't know whether the base is e or 10...
    So I googled and found some saying log(x) is normally log10(x) if not given...
    So much stuck one this three sums...
  8. Jun 2, 2012 #7
    Ok, let's say it's base a. Then [itex]t = \log_{a} (\sin x)[/itex]. When [itex]x \rightarrow \pi/2[/itex] we still have [itex]t \rightarrow 0[/itex]. When you take the antilogarithm, you have [itex]\sin x = a^{t}[/itex]. Now, express cot in terms of sin.

    Scratch that!

    As post #2 had suggested, notice the following:
    \tan x \, \log_{a}(\sin x) = \log_{a}{\left[ (\sin x)^{\tan x} \right]} = \log_{a}{\left[ (\sin x)^{\frac{1}{\cot{x}}} \right]}

    Express all the trigonometric functions in terms of cos (I will tell you why when you do it)!
    Last edited: Jun 2, 2012
  9. Jun 2, 2012 #8
    let's take a=10
    so we get the limit as ,
    lim( t->0){t/√(10-2t-1)}
    then how to proceed ?
  10. Jun 2, 2012 #9
    see my edit of the post.
  11. Jun 2, 2012 #10
    cotx= cosx/√1-cos2x
    So what ?
  12. Jun 2, 2012 #11
    So, what does the logarithm look like?
  13. Jun 2, 2012 #12
    No xp with latex sorry....I have no idea brother !
    Pls explain !
  14. Jun 2, 2012 #13
    spends a lot to type Latex...It cost my time and yours
  15. Jun 2, 2012 #14
    \log_{a} { \left[ (1 - \cos^2 x)^{\frac{\sqrt{1 - \cos^2 x}}{2 \, \cos x}} \right]}
    Now, take [itex]t = -\cos^2 x \Rightarrow \cos x = \sqrt{-t}[/itex] to get:
    \log_{a} {\left[ (1 + t)^{\frac{\sqrt{1 + t}}{2 \, \sqrt{-t}}} \right]} = \log_{a}{\left[ (1 + t)^{\frac{1}{t} \, \frac{\sqrt{-t(1 +t)}}{2}}\right]}

    Do you know the limit
    \lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}} = ?

    If you do, my work here is almost done.
  16. Jun 2, 2012 #15
    (1+0)1/0=1[itex]\infty[/itex]=[itex]\infty[/itex] ?
  17. Jun 2, 2012 #16
    (1+0)1/0=1[itex]\infty[/itex]=[itex]\infty[/itex] ?
  18. Jun 2, 2012 #17
    No. Aren't you taught any standard limits or basic formulas?
  19. Jun 2, 2012 #18
    No..only these two are available in our coursework..
    Nothing like following in ours
    \lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}
  20. Jun 2, 2012 #19
    Then, i suggest you to ask your teacher to tell you some more because you will frequently require them.
    [tex]\lim_{x\to 0} (1+x)^{\frac{1}{x}}=e[/tex]
    Solving this limit requires use logarithmic properties.

    EDIT: Woot! 700 posts.
  21. Jun 2, 2012 #20
    That's what I'm gong to do next..
    Anyway what about the other two limits ?
    [2] and [3] ?
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