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

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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` .**

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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