Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 90

For what values of $a$ and $b$ is $\displaystyle \lim_{x \to 0} \left( \frac{\sin 2x}{x^3} + a + \frac{b}{x^2} \right) = 0$
$\displaystyle \lim_{x \to 0} \left( \frac{\sin 2x}{x^3} + a + \frac{b}{x^2} \right) = 0 = \lim_{x \to 0} \frac{\sin 2x + ax^3 + bx}{x^3}$
By applying L'Hospital's Rule
$\displaystyle \lim_{x \to 0} \frac{\sin 2x + ax^3 + bx}{x^3} = \lim_{x \to 0} \frac{2\cos 2x + 3ax^2 + b}{3x^2}$

If we evaluate the limit, we will still get an indeterminate form, so we must apply L'Hospital's Rule until the limit will not result to an indeterminate form.
$\displaystyle \lim_{x \to 0} \frac{2 \cos 2x + 3ax^2 + b}{3x^2} = \lim_{x \to 0} \frac{-4\sin 2x + 6ax}{6x}$
Again, by applying L'Hospital's Rule
$\displaystyle \lim_{x \to 0} \frac{-4\sin 2x + 6ax}{6x} = \lim_{x \to 0} \frac{-4(\cos^2 x)(2) + 69}{6}$

So,

$\displaystyle \lim_{x \to 0} \frac{-4(\cos^2 x)(2) + 69}{6} = 0 $

$
\begin{equation}
\begin{aligned}
-4\cos 2(0)(2) + 6a &= 0 \\
\\
-8 \cos (0) + 6a &= 0\\
\\
-8(1) + 6a &= 0\\
\\
a &= \frac{4}{3}
\end{aligned}
\end{equation}
$


Recall that,

$
\begin{equation}
\begin{aligned}
\lim_{x \to 0} \frac{2 \cos 2x + 3ax^2 + b}{3x^2} &= 0 \\
\\
2 \cos 2x + 3ax^2 + b &= 0\\
\\
2 \cos 2(0) + 3 \left( \frac{4}{3} \right) (0)^2 + b &= 0\\
\\
2(1) + 0 + b &= 0\\
\\
b &= -2
\end{aligned}
\end{equation}
$


Therefore, we have $\displaystyle a = \frac{4}{3}$ and $b = -2$