limit x tends to 0 sin^2x/1-secx?

1 Answer
Alan P.
May 19, 2018

#lim_(xrarr0) (sin^2(x))/(1-sec(x))=color(red)(-2)#

Explanation:

#(sin^2(x))/(1-sec(x))#

#color(white)("XXX")=(1-cos^2(x))/(1-1/(cos(x))#

#color(white)("XXX")=((-1) * (cos(x)-1) * (cos(x)+1))/((cos(x)-1)/(cos(x)))#

#color(white)("XXX")=(-1) * (cos(x)+1) * cos(x)#

#color(white)("XXX")=-(cos^2(x)+cos(x))#

Note that this is defined at #x=0# [#cos^2(0)=1" and " cos(0)=1#]

So
#lim_(xrarr0) (sin^2(x))/(1-sec(x))=lim_(xrarr0)-(cos^2(x)+cos(x))#

#color(white)("XXX")=-(1+1)#

#color(white)("XXX")=-2#

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