limit
x->pi/2 \frac{x-\pi /2}{cosx}
i know by l'hopitol rule we can do....
this is my attempt without using l'hopitol rule
but
cosx=({x-\pi /2})(x-other root)..so on.....
so
\lim_{x-->\frac{\pi }{2}}(x-\frac{\pi }{2} /(x-\frac{\pi }{2})(x-root1)(x-root2)..so .on)
\lim_x-->{\frac{\pi }{2}}(\frac{1}{(x-root2)(x-root3)..()()})
which is clearly not the answer
wer is the mistake
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2 Answers
11Take x → ∩/2 + h for R.HL
lim x → ∩/2 + h lim h→0 h / (-sin h) = -1
similarily take lim x → ∩/2 - h for L.H.L
lim x→∩/2 -h lim h→0 -h / (sin h) =-1
So, L.H.L = R.H.L =-1
1708Lim (x - ∩/2) / (cos x) =
x-->∩ /2
put x - pi/2 =t
so the Limit convert into Lim (t/cos(pi/2+t)) = Lim (-t/sint) = -1
t-->0 t-->0
