How do you determine the limit of $(x-pi/2)tan(x)$ as x approaches pi/2?

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Answer 1 · 2018-05-03T17:46:21

$lim_(xrarr(pi)/2)(x-(pi)/2)tanx=-1$

$lim_(xrarr(pi)/2)(x-(pi)/2)tanx$

$(x-(pi)/2)tanx$

$=$ $(x-(pi)/2)sinx/cosx$

$(xsinx-(πsinx)/2)/cosx$

So we need to calculate this limit

$lim_(xrarrπ/2)(xsinx-(πsinx)/2)/cosx=_(DLH)^((0/0))$

$lim_(xrarrπ/2)((xsinx-(πsinx)/2)')/((cosx)'$ $=$

$-lim_(xrarrπ/2)(sinx+xcosx-(πcosx)/2)/sinx$ $=$

$-1$

because $lim_(xrarrπ/2)sinx=1$ ,

$lim_(xrarrπ/2)cosx=0$

Some graphical help enter image source here

Answer 2 · 2018-05-03T17:54:13

For an algebraic solution, please see below.

$(x-pi/2)tanx = (x-pi/2)sinx/cosx$

$= (x-pi/2)sinx/sin(pi/2-x)$

$= (-(pi/2-x))/sin(pi/2-x) sinx$

Take limit as $xrarrpi/2$ using $lim_(trarr0)t/sint = 1$ to get

$lim_(xrarrpi/2)(x-pi/2)tanx = -1$

How do you determine the limit of (x-pi/2)tan(x)(x-pi/2)tan(x) as x approaches pi/2?