Question

Question #ca2da

Answer 1

`9/2`

Explanation:

this is one i think you can hit very hard with L'Hopital's rule

again

so
`lim_(x->0)(3x-sin3x)/(x^3) = lim_(x->0)(3-3cos3x)/(3x^2) = lim_(x->0)(9sin3x)/(6x)`

but this is where we have to stop because if we re-write it as

`lim_(x->0)(9sin3x)/(6x) = lim_(x->0) 9/2 (sin3x)/(3x)`

and make the sub `p = 3x` , then we have:

`lim_(p->0) 9/2 (sin p)/(p) = 9/2 lim_(p->0) (sin p)/(p)`

we cannot use L'Hopital on that limit, because that limit is used in the derivation of `(sin x)'`

but we know from Squeeze Theorem, for example, that `lim_(p->0) (sin p)/(p) = 1`

so the limit is `9/2`