# explain inequality |sinx| <(or equal)|x| < or equal |tanx|?

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You need to remember that `tan x = sin x/cos x ` such that:

`|sin x| =lt |x| =lt |sin x|/|cos x|`

Dividing by |sin x| yields: `1 =lt |x|/|sin x| =lt 1/|cos x|`

Evaluating the limit of terms of inequality yields:

`lim_(x-gt0) 1 =ltlim_(x-gt0)|x|/|sin x| =lt lim_(x-gt0) 1/|cos x|`

`` `1=ltlim_(x-gt0)|x|/|sin x| =lt 1/(cos 0) = 1/1 = 1`

You need to use squeeze principle to confirm that limit of `|x|/|sin x|` if x->0 is 1.

**Using the squeeze pronciple it follows that `lim_(x-gt0)` `|x|/|sin x| = 1` , hence the inequality holds.**