Question

Let a&b be real numbers such that lim_"x→0"(sin3x/x^3 + a/x^2 + b)=0?

To apply l'hopitals' rule we need to make sure numerator is 0.
By doing that i can find the correct value of b to be 9/2, but i'm unable to find the correct value of a.

Answer 1

`a=-3`

`b=9/2`

Explanation:

`lim_(x->0) (sin 3x)/x^3 +a/x^2+b = lim_(x->0) (sin3x +ax+bx^3)/x^3`

The limit is in the indeterminate form `0/0` so we can apply l'Hospital's rule:

`lim_(x->0) (sin 3x)/x^3 +a/x^2+b = lim_(x->0) (d/dx (sin3x +ax+bx^3))/(d/dx x^3)`

`lim_(x->0) (sin 3x)/x^3 +a/x^2+b = lim_(x->0) (3cos 3x +a+3bx^2)/(3x^2)`

If the numerator of the last limit were non zero then the limit would be `+oo`, so we must have:

`lim_(x->0) (3cos 3x +a+3bx^2) = 0`

that is:

`3+a = 0`

`a=-3`

Thus the limit is again in the indeterminate form `0/0` and we can apply l'Hospital's rule again:

`lim_(x->0) (sin 3x)/x^3 +a/x^2+b = lim_(x->0) (d/dx (3cos 3x -3+3bx^2))/(d/dx(3x^2))`

`lim_(x->0) (sin 3x)/x^3 +a/x^2+b = lim_(x->0) (d/dx (-9sin 3x +6bx))/(d/dx(6x))`

and as this is still `0/0`, once more:

`lim_(x->0) (sin 3x)/x^3 +a/x^2+b = lim_(x->0) (-27cos 3x +6b)/6 =-9/2 +b`

Thus we must have:

`a=-3`

`b=9/2`

graph{(sin (3x))/x^3 -3/x^2+9/2 [-10, 10, -5, 5]}