# 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

(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:

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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:

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