Mathematics

Problem

Which of the following statements best justifies that

\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1

? a) Since for very small angles (in radians)

\sin x\approx x

, replacing

\sin x

x

gives

\dfrac{x}{x}=1

, so the limit is

1

. b) Because

\sin x

is continuous at

x=0

and

\sin 0=0

, the ratio

\dfrac{\sin x}{x}

is also continuous there, hence equals

\dfrac{0}{0}=1

. c) Because

\sin x

is differentiable at

x=0

and

\displaystyle \frac{d}{dx}(\sin x)\big|_{x=0}=1

. d) Applying the limit law

\displaystyle \frac{\lim_{x\to0}\sin x}{\lim_{x\to0}x}=\frac{0}{0}

and canceling the zeros leaves

1

, so the limit is

1