Evaluate the limit-x -2limsin 2xcos x

We have, x →π2limsin 2xcos x Put x=π2+h, h → 0 =h → 0limsin 2 ( π2+h )cos nbsp;( π2+h ) =h → 0limsin( π+2h )cos nbsp;( π2+h ) [ ∵sin (π+θ)= sinθ, c ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

Evaluate the ...

Question

Evaluate the limit:
$lim x \to \frac{π}{2} \frac{sin 2 x}{cos x}$

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Solution

We have,

$lim x \to \frac{π}{2} \frac{sin 2 x}{cos x}$

Put $x = \frac{π}{2} + h, h \to 0$

$= lim h \to 0 \frac{sin 2 (\frac{π}{2} + h)}{cos (\frac{π}{2} + h)}$

$= lim h \to 0 \frac{sin (π + 2 h)}{cos (\frac{π}{2} + h)}$

$[∵ sin (π + θ) = - sin θ, cos (\frac{π}{2} + h) = - sin θ]$

$= lim h \to 0 \frac{- sin 2 h}{- sin h}$

$= lim h \to 0 \frac{2 sin h cos h}{sin h}$ $[∵ sin 2 θ = 2 sin θ cos θ]$

$= lim h \to 0 2 cos h$

$= 2 (1) = 2$

Therefore,

$lim x \to \frac{π}{2} \frac{sin 2 x}{cos x} = 2$

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General Solution of Trigonometric Equation

Standard XII Mathematics

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Evaluate the limit: limx→π2sin2xcosx

We have, limx→π2sin2xcosx Put x=π2+h,h→0 =limh→0sin2(π2+h)cos(π2+h) =limh→0sin(π+2h)cos(π2+h) [∵sin(π+θ)=−sinθ,cos(π2+h)=−sinθ] =limh→0−sin2h−sinh =limh→02sinhcoshsinh [∵sin2θ=2sinθcosθ] =limh→02cosh =2(1)=2 Therefore, limx→π2sin2xcosx=2

Evaluate the limit:
$lim x \to \frac{π}{2} \frac{sin 2 x}{cos x}$