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Post by Scott Beck on April 24 at 09:15:09 AM

Shouldn't the pre-set limit rule be (1-cos x)/x rather than what you have?

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Post by Samiha Bushra on February 18 at 08:06:14 PM

I don't understand how sinx/x is equivalent to 1. sin(0) is 0 and if x is 0, shouldn't it be 0 because anything divided by 0 is 0?

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Post by GUOJI LIU on November 6, 2012

take look the graph f(x)=cos(x), you will find out who when x approaches to 0, the value of function will be 1

1 answer

Last reply by: GUOJI LIU
Tue Nov 6, 2012 10:50 AM

Post by Kirby Baker on September 27, 2012

how is cos(x) equail to 1

Special Trigonometric Limits

  • Reorder terms to resemble special trig limit forms
  • Useful identity:

Special Trigonometric Limits

Find limx → 0 [1/x] (x − [(x3)/6] + [(x5)/120] − [(x7)/5040])

  • limx → 0 [1/x] (x − [(x3)/6] + [(x5)/120] − [(x7)/5040]) = limx → 0 ([x/x] − [(x3)/6x] + [(x5)/120x] − [(x7)/5040x])
    = 1 − 0 + 0 − 0 = 1
  • The expression on the right of the original limit just happens to be an approximation of the series representation of sin(x). This representation is accurate to within .0003% for −1 < x < 1.
  • sin(x) = x − [(x3)/3!] + [(x5)/5!] − [(x7)/7!] + ... for all x
  • Reference: the exclamation point is the factorial symbol. Roughly speaking, it equates to one times every integer less than or equal to that number that is still greater than zero. 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1

1

Find limx → 0 [sin(2x)/x]

  • limx → 0 [sin(2x)/x] = limx → 0 [1/x] (2x − [((2x)3)/6] + [((2x)5)/120] − [((2x)7)/5040] + ...)
  • = limx → 0 ([2x/x] − [((2x)3)/6x] + [((2x)5)/120x] − [((2x)7)/5040x] + ...)
  • = limx → 0 [2x/x] + 0
  • = 2 limx → 0 [x/x] = 2 * 1 = 2
  • In general, limx → 0 [sin(ax)/x] = a

2

Find limx → 0 [sin(9x)/sin(7x)]

  • limx → 0 [sin(9x)/sin(7x)] = limx → 0 [sin(9x)/sin(7x)] [x/x]
  • = limx → 0 [sin(9x)/x] [x/sin(7x)]
  • = limx → 0 [sin(9x)/x] limx → 0 [x/sin(7x)]
  • = 9 [1/7] = [9/7]
  • In general, limx → 0 [sin(ax)/sin(bx)] = [a/b]

[9/7]

Find limx → 0 [1/x] ((1 − [(x2)/2] + [(x4)/12] − [(x6)/720]) − 1)

  • limx → 0 [1/x] ((1 − [(x2)/2] + [(x4)/12] − [(x6)/720]) − 1) = limx → 0 [1/x] (− [(x2)/2] + [(x4)/12] − [(x6)/720])
  • = limx → 0 (− [(x2)/2x] + [(x4)/12x] − [(x6)/720x])
  • = 0
  • The bigger exponents in the numerators dominate. In the inner parenthesis, an approximation of the series representation of cos(x) can be seen.
  • cos(x) = 1 − [(x2)/2!] + [(x4)/4!] − ... for all x

= 0

Find limx → 0 [(cos2 x − 1)/13x sin x]

  • limx → 0 [(cos2 x − 1)/13x sin x] = [1/13] limx → 0 [(cos2 x − 1)/x sin x]
  • = [1/13] limx → 0 [(−sin2 x)/x sin x]
  • = −[1/13] limx → 0 [sin x/x] = −[1/13]

−[1/13]

Find limx → 0 [sin(2x)/x cos(x)]

  • limx → 0 [sin(2x)/x cos(x)] = limx → 0 [sin(2x)/x] limx → 0 [1/cos(x)]
  • = 2 * 1 = 2

2

Find limx → 0 [sin(2x)/cos(x)sin(3x)]

  • limx → 0 [sin(2x)/cos(x)sin(3x)] = limx → 0 [2sin(x)cos(x)/cos(x)sin(3x)]
  • = 2 limx → 0 [sin(x)/sin(3x)]
  • = [2/3]

= [2/3]

Find limx → 0 [(sin(2x) − 2sin(x))/x sin(x)]

  • limx → 0 [(sin(2x) − 2sin(x))/x sin(x)] = limx → 0 [(2sin(x)cos(x) − 2sin(x))/x sin(x)]
  • = limx → 0 [(2cos(x) − 2)/x]
  • = 2 limx → 0 [(cos(x) − 1)/x]
  • = 2 * 0 = 0

0

Find limx → 0 [(5cos(x)sin(x) − 5sin(x))/(x2)]

  • limx → 0 [(5cos(x)sin(x) − 5sin(x))/(x2)] = limx → 0 [(5sin(x)(cos(x) − 1))/(x2)]
  • = 5 limx → 0 [sin(x)/x] limx → 0 [(cos(x) − 1)/x]
  • = 5 * 1 * 0 = 0

0

Find limx → 0 [tan(5x)/x]

  • limx → 0 [tan(5x)/x] = limx → 0 [sin(5x)/x cos(5x)]
  • = limx → 0 [sin(5x)/x] limx → 0 [1/cos(5x)]
  • = 5 * 1 = 5

5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Special Trigonometric Limits

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