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Homework Help: Limits problem

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [tex]f(x)=\frac{sin^{-1}(1-\{x\})\cdot cos^{-1}(1-\{x\})}{\sqrt{2\{x\}}\cdot (1-\{x\})}[/tex] then find [itex]\lim_{x→0^+}f(x)[/itex] and [itex]\lim_{x→0^-}f(x)[/itex], where {x} denotes the fractional part function.

    2. Relevant equations

    3. The attempt at a solution
    I have solved [itex]\lim_{x→0^-}f(x)[/itex], using [itex]\lim_{g(x)→0} \frac{sin^{-1}g(x)}{g(x)}=1[/itex]. If we approach a fractional part function at 0 from left, we get the value as 1. Therefore i get my answer to be [itex]\frac{\pi}{2\sqrt{2}}[/itex]/

    I am stuck for the first part, [itex]\lim_{x→0^+}f(x)[/itex]. When we approach the fractional part function at 0 from right, its value becomes zero. Due to this i get a 0/0 form.
    I am not allowed to use L'Hôpital's rule.

    Any help is appreciated. :smile:
    Last edited: Jun 2, 2012
  2. jcsd
  3. Jun 2, 2012 #2
    Solving this without L'Hospital's is fun! Here's how I approached it,

    Since [itex]x\to 0^+[/itex], the fractional part of x, i.e [itex]\left \{x \right \}[/itex] will behave as [itex]x[/itex].

    This gives you the equation as,

    [tex]\frac{sin^{-1}(1-x)\cdot cos^{-1}(1-x)}{\sqrt{2x}\cdot (1-x)}[/tex]


    [tex]\lim_{x\to 0^+}\frac{sin^{-1}(1-x)}{(1-x)} \cdot \lim_{x\to 0^+} \frac{cos^{-1}(1-x)}{\sqrt{2x}}[/tex]

    The limit of the first part is trival, and comes out to be [itex]\pi/2[/itex]. The second part is the one that is confusing(without L'Hospital's). Can you try it out?
  4. Jun 2, 2012 #3
    Thanks Infinitum! I too was stuck at the same point. :smile:
    I have figured it out, i solved the second part and it came out be one.
    Here are the steps:
    [tex]\lim_{x\to 0^+} \frac{cos^{-1}(1-x)}{\sqrt{2x}}=\lim_{x\to 0^+} \frac{sin^{-1}\sqrt{2x-x^2}}{\sqrt{2x}}[/tex]

    Using [itex]\lim_{g(x)→0} \frac{sin^{-1}g(x)}{g(x)}=1[/itex],
    the solution of limit is 1 and hence the answer is [itex]\frac{\pi}{2}[/itex].

    Thanks once again. :smile:
  5. Jun 2, 2012 #4
    Yep!! That's correct! :approve:

    To me, that first step transformation was the most troublesome, glad you figured it out!
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