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Can someone help me show that these inequalities are true:

$|\sinh x| ≤ 3 |x| , |\cosh x-1|≤ 3|x|$ for $|x| < 1/2 $

I also have to show that they are continuous

I'm not allowed to use calculus

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    Use Taylor polynomials and the remainder to set up the maximal error in the approximation. – 2017-02-16
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    Use Lagrange's MVT. – 2017-02-16
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    It's pretty rude to edit questions like that,check your questions before you post them and please refrain from drastically changing the question and adding new conditions especially when there already are answers. – 2017-02-16

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$|\sin x| \le |x|$ so certainly $|\sin x| \le 3|x|$

$|\cosh x - 1|\cdots$ I don't have anything quite so elegant except to say.

$\cosh x - 1$ is an even function. $\cosh x-1$ is monotonically increasing when $x>0$

You only need to test the endpoint $x= \frac 12$.

I see an update.... you meant to say $|\sinh x| \le 3|x|$

$|\tanh x| \le |x|\\ |\sinh x|\le |x \cosh x|$

when $-\frac 12

$|\sinh x|\le 3|x|$

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    I'm not allowed to differentiate, and I accidently typed sin instead of sinh :( – 2017-02-16