# 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.

Feb 23, 2018

$a = - 3$

$b = \frac{9}{2}$

#### Explanation:

${\lim}_{x \to 0} \frac{\sin 3 x}{x} ^ 3 + \frac{a}{x} ^ 2 + b = {\lim}_{x \to 0} \frac{\sin 3 x + a x + b {x}^{3}}{x} ^ 3$

The limit is in the indeterminate form $\frac{0}{0}$ so we can apply l'Hospital's rule:

${\lim}_{x \to 0} \frac{\sin 3 x}{x} ^ 3 + \frac{a}{x} ^ 2 + b = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(\sin 3 x + a x + b {x}^{3}\right)}{\frac{d}{\mathrm{dx}} {x}^{3}}$

${\lim}_{x \to 0} \frac{\sin 3 x}{x} ^ 3 + \frac{a}{x} ^ 2 + b = {\lim}_{x \to 0} \frac{3 \cos 3 x + a + 3 b {x}^{2}}{3 {x}^{2}}$

If the numerator of the last limit were non zero then the limit would be $+ \infty$, so we must have:

${\lim}_{x \to 0} \left(3 \cos 3 x + a + 3 b {x}^{2}\right) = 0$

that is:

$3 + a = 0$

$a = - 3$

Thus the limit is again in the indeterminate form $\frac{0}{0}$ and we can apply l'Hospital's rule again:

${\lim}_{x \to 0} \frac{\sin 3 x}{x} ^ 3 + \frac{a}{x} ^ 2 + b = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(3 \cos 3 x - 3 + 3 b {x}^{2}\right)}{\frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right)}$

${\lim}_{x \to 0} \frac{\sin 3 x}{x} ^ 3 + \frac{a}{x} ^ 2 + b = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(- 9 \sin 3 x + 6 b x\right)}{\frac{d}{\mathrm{dx}} \left(6 x\right)}$

and as this is still $\frac{0}{0}$, once more:

${\lim}_{x \to 0} \frac{\sin 3 x}{x} ^ 3 + \frac{a}{x} ^ 2 + b = {\lim}_{x \to 0} \frac{- 27 \cos 3 x + 6 b}{6} = - \frac{9}{2} + b$

Thus we must have:

$a = - 3$

$b = \frac{9}{2}$

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