Question

Prove that lim_(x→0) sinx/x = 1 Where “x” being measured in radians.

Answer 1

lim_(θ→0) sinθ/θ = 1

Proof: Consider a circle with centre ‘O’ and radius ‘r’. Mark two point A and B on the circumference of the circle so that (AOB)! = θ

At ‘A’ draw a tangent to the circle produce OB to cut the tangent at C. 

Joint AB. 

Draw BM ⊥ OA, 

Here OA = OB = r 

From the figure

Area of triangle OAB < area of the sector AOB < area of triangle OAC

Answer 2

L.H.S. = \(\lim\limits_{x\to 0} \frac {\sin x}x\)

\( = \frac{\sin (0)}0\)      \(\left(\frac 00\,\text{form}\right)\)

L-Hospital Rule

\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}\)

\(\lim\limits_{x\to 0} \frac{\sin x}x = \lim\limits_{x\to 0}\frac{\cos x}1\)

\(= \frac{\cos (0)}1\)

\(= \frac11\)

\(= 1\)

= R.H.S.

Hence proved.