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

sin3x=sin(2x+x)=sin(2x)cos(x)+sin(x)cos(2x)

=`2sin(x)cos^2(x)+2sin(x)cos^2(x)-sin(x)`

`=4sin(x)cos^2(x)-sin(x)`

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

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

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

`=4(1)(1)-1=3`

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

The graph:

` `

` `

`lim_(x->0)sin(x)/x=1`

`Thus`

`lim_(x->0)sin(3x)/x=lim_(x->0)(3sin(3x))/(3x)`

`=3lim_(3x->0)sin(3x)/(3x)=3`

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