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Q. `lim_(x->pi/2)` `(pi/2 - x)tanx` .Evaluate.

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user8235304 | Student, Grade 11 | (Level 1) Valedictorian

Posted June 29, 2013 at 12:39 AM via web

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Q. `lim_(x->pi/2)` `(pi/2 - x)tanx` .Evaluate.

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mjripalda | High School Teacher | (Level 1) Senior Educator

Posted June 29, 2013 at 1:37 AM (Answer #1)

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`lim_(x->pi/2) (pi/2-x)tanx`

Note when x=pi/2 is plug-in to the function, the result is indeterminate.

`lim_(x->pi/2) (pi/2-x)tanx=(pi/2-pi/2)tanpi/2=0*oo`

So, to determine its limit, L'Hospital's rule can be applied.

To do so, express tan x as ratio of sine and cosine.

`lim_(x->pi/2) (pi/2-x)tanx ` 

`=lim_(x->pi/2)(pi/2-x)*sinx/cosx`

`=lim_(x->pi/2) ((pi/2-x)sinx)/cosx`         

Then, take the derivative of the numerator and denominator of the function separately.

`=lim_(x->pi/2) (((pi/2-x)sinx)')/((cosx)')`                   

`=lim_(x->pi/2) ((pi/2-x)*(sinx)' + sinx* (pi/2-x)')/((cosx)')`            

`=lim_(x->pi/2) ((pi/2-x)cosx + sinx(-1))/(-sinx)`  

`=lim_(x->pi/2) ((pi/2-x)cosx-sinx)/(-sinx)`      

And then, plug-in x=pi/2.

`= ((pi/2-pi/2)cos(pi/2)-sin (pi/2))/(-sin (pi/2))`

Since  `cos(pi/2)=0`  and  `sin(pi/2)=1` , it simplifies to:

`=(0*0-1)/(-1)`

`=(-1)/(-1)`

`=1`

Hence,  `lim_(x->pi/2) (pi/2-x)tanx=1` .

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aruv | High School Teacher | (Level 2) Valedictorian

Posted June 30, 2013 at 9:01 AM (Answer #2)

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It is

`lim_(x->pi/2)(pi/2-x)tan(x)`

`=lim_{x->pi/2}{(pi/2-x)sin(x)}/(cos(x))`

`=lim_{x->pi/2}{(pi/2-x)sin(x)}/(sin(pi/2-x))`

`lim_{x->pi/2}(pi/2-x)/sin(pi/2-x)=lim_{x->pi/2}{(-(x-pi/2))/(-sin(x-pi/2))}=1`

`since`

`lim_{x->0}sin(x)/x=1`

`therefore `

`lim_{x->pi/2}{(pi/2-x)/sin(pi/2-x)}lim_{x->pi/2}sin(x)=1`

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