Question

Question #1f9ed

Answer 1

`lim_(xrarr0)(xcosx+x)/(sinxcosx )=2`

Explanation:

Given `lim_(xrarr0)(xcosx+x)/(sinxcosx )` first factor the numerator:

`lim_(xrarr0)(x(cosx+1))/(sinxcosx )`

Now rearrange a little to make things more explicit:

`lim_(xrarr0)(x/sinx*(cosx+1)/cosx)`

as long as both limits exist we can change this to a product of two limits:

`lim_(xrarr0)(x/sinx)*lim_(xrarr0)((cosx+1)/cosx)`

Since `lim_(xrarr0)(sinx/x)=1` its reciprocal will also be 1, that is `lim_(xrarr0)(x/sinx)=1`, so:

`lim_(xrarr0)(x/sinx)*lim_(xrarr0)((cosx+1)/cosx)=1*lim_(xrarr0)((cosx+1)/cosx)`

The remaining limit can be evaluated by direct substitution:

`lim_(xrarr0)((cosx+1)/cosx)=(1+1)/1=2`

Answer 2

`lim_(x->0)(xcosx+x)/(sinxcosx )=2`

Explanation:

`lim_(x->0)(xcosx+x)/(sinxcosx )`

`-=lim_(x->0)(xcosx)/(sinxcosx )+lim_(x->0)(x)/(sinxcosx )`

`-=lim_(x->0)(x)/(sinx)+lim_(x->0)(x)/(sinxcosx)`

`-=lim_(x->0)(x)/(sinx)+lim_(x->0)(x)/(sinx)*lim_(x->0)1/cosx`

Since `lim_(x->0)(sinx)/(x )=1` then `lim_(xrarr0)x/sinx=1`

`-=1+1*lim_(x->0)1/cosx`

`-=1+lim_(x->0)1/cosx`

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

`-=1+1=2`

`lim_(x->0)(xcosx+x)/(sinxcosx )=2`