Use theorem: sinx/x =1  and cos ((x-1)/x) = 0 as x approaches 0 to solve the following: a.  sin 3x/x  as x approaches 0 b. tan 5x/x  as x approaches 0

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embizze's profile pic

Posted on

(a) Find `lim_(x->0)(sin3x)/x` :




Then `lim_(x->0)(sin3x)/x=lim_(x->0)(4sinxcos^2x-sinx)/x`


`=lim_(x->0)4*lim_(x->0)(sinx)/x * lim_(x->0)cos^2x-1`


So `lim_(x->0)(sin3x)/x=3`

The graph:



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aruv's profile pic

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