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

Question

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.

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Answer 1 · 2018-02-23T00:05:58

#a=-3#

#b=9/2#

#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]}