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Indeterminate limit problem

  1. Jun 1, 2014 #1
    1. The problem statement, all variables and given/known data
    i need to find the value of limit but its indeterminate and i cant use l'hospital's rule.
    i multiplied nominator and denominator with [itex]\displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}{|x|}}[/itex] to get rid of square root. After get i tried to make some transforms to simplify something but i couldn find anything right.


    2. Relevant equations
    This is the question:
    [itex]
    \displaystyle\lim_{x\rightarrow 0^+} {\frac{\sqrt{2sin^{2}(x/2)+1-cos^2(x)}}{|x|}}
    [/itex]


    3. The attempt at a solution
    this is the last step that i came.
    [itex]
    \displaystyle\lim_{x\rightarrow 0^+} {\frac{1-cos(x)+1-cos^{2}(x)}{|x|\sqrt{2sin^{2}(x/2)+1+cos^2(x)}}}
    [/itex]
     
  2. jcsd
  3. Jun 1, 2014 #2

    pasmith

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    Does recognising that if [itex]f(0) = 0[/itex] then [tex]
    f'(0) = \lim_{x \to 0^{+}} \frac{f(x)-f(0)}{x} = \lim_{x \to 0^{+}} \frac{f(x)}{|x|}
    [/tex]count as "using l'Hopital's rule"?

    I personally would use [itex]\cos^2(x/2) - \sin^2(x/2) = \cos \theta[/itex] to express [itex]2\sin^{2}(x/2)+1+\cos^2(x)[/itex] as a quadratic in [itex]\cos x[/itex] and then complete the square before differentiating.
     
  4. Jun 1, 2014 #3

    lurflurf

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    $$\frac{\sqrt{2\sin^{2}(x/2)+1-\cos^2(x)}}{|x|}=\sqrt{\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2+\left(\frac{sin(x)}{x}\right)^2}$$
    Do you know what
    $$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=\sin^\prime(0)$$
    is?

    I would say recognizing
    $$\lim_{x\rightarrow 0}\frac{\mathrm{f}(x+a)-\mathrm{f}(a)}{x}=\mathrm{f}^\prime(a)$$
    does not count as "using l'Hopital's rule."
     
  5. Jun 1, 2014 #4

    lurflurf

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    you have a mistake somewhere post more steps if you cannot find it yourself

    you also could use
    $$2\sin^2(x/2)+1-\cos^2(x)=2\sin^2(x/2)(2+\cos(x))$$
     
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