Question #3ade9

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Answer 1 · 2017-04-14T12:13:17

#-2#

I edited your question a bit, and I hope that this is the limit you are looking for.

By lH#hat"o"#pital's Rule,

#lim_(theta to pi/2)(2theta-pi)/(cos theta)=lim_(theta to pi/2)2/-sin theta=2/(-sin(pi/2))=2/(-1)=-2#

I hope that this was clear.

Answer 2 · 2017-04-14T12:30:52

#-2.#

Without using L'Hospital's Rule, the Limit can be derived as

shown below:

Let, #theta=x+pi/2. :. theta to pi/2 rArr x to 0.#

#:." The Reqd. Lim.="lim_(x to 0){2(x+pi/2)-pi}/cos(x+pi/2),#

#=lim_(x to 0) (2x)/(-sinx)=(-2){lim_(x to 0)x/sinx}.#

#:." The Reqd. Lim."=-2.#

Enjoy Maths.!