Question

prove geometrically lim x→0 sinx/x=1 where x is the radian and hence evaluate limx→0 sin4x/sin2x​

Answer 1

Answer:

1...Proof:- Consider a circle with centre 0 & radius 'r' Mark two points A & B on the circle so that

∠AOB = θ radians.

At A draw a tangent to the circle.

Produce OB to cut the tangent at C. Join AB. Draw BM ꓕ OA.

From the figure it is clear that Area of AOAB < Area of sector OAB <sector of ∆OAC.... (1)

Area of ∆OAB=OA.BM

[In ∆OBM.sinθ=BM\OB=BM/r ⟹ BM=r sinθ]

∴ Area of OAB=1/2rr.sinθ

=1/2r²θ

Area of sector OAB=1/2r²θ

Area of ∆OAC=1/2OA.OC

=1/2r²tanθ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎

‎‎ ‎[In ∆OAC,tanθ= AC/OA=AC/r=tan θ]

∴ (1 )becomes

1/2r²sinθ < 1/2r²θ < 1/2r²tanθ

devide by 1/2r²

⟹sinθ < θ < tanθ ...(2)

devide 2 by sin θ;

‎‎ ‎‎ sinθ/sinθ < θ/sinθ <tanθ/sinθ

⟹1 < sinθ/θ < 1/cosθ

taking reciprocal;

1 <sinθ/θ < cosθ

Taking the limit as θ→0

lim θ→0 >lim θ→0 ‎‎.sinθ/θ > lim θ→0 ‎‎.cosθ

‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎ ‎‎

⟹1>limθ→0 sinθ/θ >1

⟹limθ→0 sinθ/θ =1

___________________________

2...and hence evaluate limx→0 sin4x/sin2x

refer the attachment (image 2)