Question

How do solve without using derivative?

`lim xrarr0 (sin(x^3+x^2-x)+sinx)/x`

Answer 1

`lim_(x->0)frac (sin(x^3+x^2-x)+sinx) x =0`

Explanation:

You can resolve this limit without using derivatives by re-conducing it to the known limit:

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

We can proceed in this way:

`frac (sin(x^3+x^2-x)+sinx) x = frac (sin(x^3+x^2-x) )x + frac (sinx) x = frac (sin(x^3+x^2-x) ) (x^3+x^2-x)* frac (x^3+x^2-x) x+ frac (sinx) x = frac (sin(x^3+x^2-x) ) (x^3+x^2-x)* (x^2+x-1) + frac (sinx) x`

Thus:
`lim_(x->0)frac (sin(x^3+x^2-x)+sinx) x =lim_(x->0)frac (sin(x^3+x^2-x) ) (x^3+x^2-x)* (x^2+x-1) + frac (sinx) x = 1*(-1)+1=0`