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Evaluate````` lim_(x->0) (1-cos^2(2sinx))/(1-cos2x)` ` ` `` ``

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roshan-rox | (Level 1) Valedictorian

Posted June 4, 2013 at 4:35 PM via web

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Evaluate````` lim_(x->0) (1-cos^2(2sinx))/(1-cos2x)`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted June 4, 2013 at 5:45 PM (Answer #2)

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`lim_(xrarr0)(1-cos^2(2sinx))/(1-cos2x)`

We know that;

`sin^2x = 1-cos^2x`

`cos2x = 1-2sin^2x`

 

`lim_(xrarr0)(1-cos^2(2sinx))/(1-cos2x)`

`= lim_(xrarr0)(sin^2(2sinx))/(1-(1-2sin^2x))`

`= lim_(xrarr0)(2sin^2(2sinx))/(2sinx)^2`

 

Let;

`sinx = t`

When `xrarr 0` then `sinx rarr 0` so `t rarr 0`

 

`= lim_(xrarr0)(2sin^2(2sinx))/(2sinx)^2`

`= 2lim_(trarr0)((sint)/t)^2`

 

We know that `lim_(xrarr0)(sinx)/x = 1`

So `lim_(trarr0)((sint)/t)^2 = 1^2 = 1`

 

`= lim_(xrarr0)(2sin^2(2sinx))/(2sinx)^2`

`= 2lim_(trarr0)((sint)/t)^2`

`= 2xx1`

`= 2`

 

So the answer is;

`lim_(xrarr0)(1-cos^2(2sinx))/(1-cos2x) = 2`

 

Sources:

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted June 5, 2013 at 7:10 AM (Answer #3)

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Given  problem is  `lim_(x->0)(1-cos^2(2sinx))/(1-cos(2x))` 

Now we know that  `1-cos^2x=sin^2x`  and   `cos2x=1-2sin^2x` . Substituting these values inthe above expression we get

            =`lim_(x->0)(sin^2(2sinx))/(1-1+2sin^2x)`

           =`lim_(x->0) (sin^2(2sinx))/(2sin^2x)`

  Let  `2sinx=t` , now as `x->0, t->0`

so,    = `lim_(t->0)(sin^2t)/(t^2/2)`

         =`2lim_(t->0)(sint/t)^2`

        =`2(lim_(t->0)sint/t)^2`  .      As `lim_(t->0)sint/t=1`

       =2 .          

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 4, 2013 at 4:48 PM (Answer #1)

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You need to use the following definition of limit such that:

`lim_(x->a) f(x) = f(a)`

Reasoning by analogy yields:

`lim_(x->0) (1 - cos^2(2 sin x))/(1 - cos 2x) = (1 - cos^2(2 sin 0))/(1 - cos 0)`

Since `sin 0 = 0` and `cos 0 = 1` yields:

`lim_(x->0) (1 - cos^2(2 sin x))/(1 - cos 2x) = (1 -1)/(1 -1) = 0/0`

The indetermination case ` 0/0` requests for you to use l'Hospital's theorem such that:

`lim_(x->0) (1 - cos^2(2 sin x))/(1 - cos 2x) = lim_(x->0) ((1 - cos^2(2 sin x))')/((1 - cos 2x)') `

`lim_(x->0) ((1 - cos^2(2 sin x))')/((1 - cos 2x)') = lim_(x->0) (4cos(2sin x)*sin(2sin x)*cos x)/(2sin 2x) `

Replacing `2 sin x cos x` for `sin 2x` yields:

`lim_(x->0) (4cos(2sin x)*sin(2sin x)*cos x)/(2sin 2x)= lim_(x->0) (4cos(2sin x)*sin(2sin x)*cos x)/(4sin x*cos x)`

Reducing duplicate factors yields:

`lim_(x->0) (cos(2sin x)*sin(2sin x))/(sin x) = 0/0`

Using again l'Hospital's theorem yields:

`lim_(x->0) (sin(4sin x))/(2sin x) = lim_(x->0) ((sin(4sin x))')/((2sin x)')`

`lim_(x->0) (4cos(4sin x)*cos x)/(2cos x)`

Reducing duplicate terms yields:

`lim_(x->0) (2cos(4sin x))/1 = 2`

Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->0) (1 - cos^2(2 sin x))/(1 - cos 2x) = 2.`

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