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



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Top Answer

mjripalda's profile pic

Posted on (Answer #1)

`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`         

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:




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

Top Answer

aruv's profile pic

Posted on (Answer #2)

It is







`therefore `





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