How do you find the limit of #sec3xcos5x# as x approaches pi/2 from the left?

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Answer 1 · 2016-08-20T18:34:53

Reqd. Limit#=-5/3#.

Reqd. Limit#=lim_(xrarrpi/2-) sec3xcos5x#

Since #xrarrpi/2-, x<,pi/2#. Let #x+h=pi/2#

#:. x rarr pi/2- rArr h rarr 0+#

Now, #sec3xcos5x=sec3(pi/2-h)cos5(pi/2-h)#

#=sec(3pi/2-3h)cos(5pi/2-5h)=(-csc3h)(sin5h)=-sin(5h)/sin(3h)#

#=-[{(sin(5h)/(5h))*5h}/{(sin(3h)/(3h))*3h}]#

#=-5/3[{(sin(5h)/(5h))}/{(sin(3h)/(3h))}]#

Since, #lim_(hrarr0+) sin(5h)/(5h)=lim_(hrarr0+) sin(3h)/(3h)=1#,

Reqd. Limit#=-5/3#.

Enjoy Maths.!