Question

How do you find the limit of `[1/(3+X)]- (1/3) ÷ X` as x->0?

Answer 1

Assuming you meant:

`lim_(xrarr0)((1/(3+x)-1/3)-:x)=lim_(xrarr0)(1/(3+x)-1/3)/x`

Multiply the fraction by the inner denominators:

`=lim_(xrarr0)(1/(3+x)-1/3)/x*((3+x)(3))/((3+x)(3))=lim_(xrarr0)(3-(3+x))/(x(3+x)(3))`

`=lim_(xrarr0)(-x)/(3x(x+3))=lim_(xrarr0)(-1)/(3(x+3))`

Now we can evaluate the limit:

`=(-1)/(3(0+3))=-1/9`