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.

1 Answer
Feb 23, 2018

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]}