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Limit of tan(x)/x as x approaching zero

  1. Apr 5, 2015 #1
  2. jcsd
  3. Apr 5, 2015 #2

    Svein

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    Hints: [itex]tan(x)=\frac{sin(x)}{cos(x)} [/itex] and cos(0) = 1.
     
  4. Apr 5, 2015 #3
    Furthermore use sqeeze theorem to calculate [tex]\lim_{x\to 0}{\frac{\sin{x}}{x}}[/tex]
    L'Hospital rule will not work in that limit
     
  5. Apr 5, 2015 #4

    Svein

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    Why not?
     
  6. Apr 5, 2015 #5
    I know that I can make it lim x->0 tan(x) / x = lim x->0 1/cos(x) * sin(x)/x = 1 * 1 = 1

    But, the hint in http://math.stackexchange.com/quest...-lim-limits-x-to0-frac-tan-xx-1#answer-448210 says

    lim x->0 tan(x)/x = lim x->0 tan(x)-tan(0)/ x-0
    If I plug x=0, the denominator will be zero.
    Then, how to solve it using that hint ?
    What trigonometric identity should be used ?
     
  7. Apr 5, 2015 #6

    Svein

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    Well, using l'Hôpital directly, you get...
     
  8. Apr 5, 2015 #7

    Mark44

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    Sure it will.
    ##\lim_{x \to 0}\frac{sin(x)}{x} = \lim_{x \to 0}\frac{cos(x)}{1} = 1##
     
  9. Apr 5, 2015 #8

    pasmith

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    L'Hopital's rule does indeed work here, with the caveat that the argument is circular if you are trying to prove that [itex]\sin'(0) = \lim_{x \to 0} \frac{\sin x}x[/itex] exists and is equal to 1.
     
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