- #1

Benny

- 584

- 0

[tex]

\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1

[/tex]

I think that it can be done using the squeeze theorem but I can't get to the answer. I started with [tex] - 1 \le \sin x \le 1[/tex]. At this point dividing through by x gets me nowhere since the limits won't be equal. So I try to divide by cos(x) which gives: [tex] - \sec x \le \tan x \le \sec x[/tex].

The left part of the inequality approaches negative one as x approaches zero and the right part of the inequality approaches positive one as x approaches zero. I can't go any further so can someone please help me out?