If the limit $\lim\limits_{x\to 0}{\frac{\sin 3x}{x^3} + \frac{a}{x^2}+b}$ exists and equals $0$ then what can $a$ and $b$ be?

Question

Let $$L=\lim\limits_{x\to 0}{\frac{\sin 3x}{x^3} + \frac{a}{x^2}+b}=0$$ given that $a,b \in \mathbb R$ and are finite.

I tried the following approach,

We know, $\lim\limits_{x\to 0}{\frac{\sin 3x}{3x}}=1$

$$\therefore L=\lim\limits_{x\to 0}{\frac{\sin 3x}{3x}\cdot \frac{3}{x^2}+\frac{a}{x^2}+b}$$

substituting $\lim\limits_{x\to 0}{\frac{\sin 3x}{3x}}=1$

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

after some simplification $$L=\lim\limits_{x\to 0}{\frac{bx^2+a+3}{x^2}}$$

Now, for L to exist $bx^2+a+3 \to 0$ Hence, $$\boxed{a=-3}$$

and as $$L=0 \implies \boxed{b=0}$$ but this is wrong! $b \not= 0$

I tried to plot $f(x)=\frac{\sin 3x}{x^3}-\frac{3}{x^2}$ (image attached here). Clearly from graph $b=\frac{9}{2}$. But what is wrong with this approach?

Answer 1

In a sum, you cannot apply $\sin(x)\approx x$.

The correct approach would be first looking for a common denominator: $$\dfrac{bx^3+ax+\sin(3x)}{x^3}$$

By Taylor's Theorem, we know $$\sin(3x)=3x-\dfrac{27}{6}x^3+g(x)x^4=3x-\dfrac{9}{2}x^3+g(x)x^4$$

for some function $g$ such that $g(x)\to 0$ when $x\to 0$.

Thus, the expression transforms to $$\dfrac{(b-9/2)x^3+(a+3)x+g(x)x^4}{x^3}$$

For its limit to exist we need $$a+3=0\iff \boxed{a=-3}$$

And we will have $$\lim_{x\to 0} \dfrac{bx^3+ax+\sin(3x)}{x^3}=b-\dfrac{9}{2}$$ for any $b\in\mathbb{R}$. Thus, we will need $b=\dfrac{9}{2}$ for $L$ to be $0$.

Answer 2

You cannot just "substitute" $\lim\limits_{x \to 0} \frac{\sin(3x)}{3x} = 1$. A proper way of doing it is using Landau notation, i.e., $$\frac{\sin(3x)}{3x} = 1 - \frac{3}{2}x^2 + O(x^4).$$ Then you get $$L = \lim\limits_{x\to 0} \frac{bx^2+a+3(1 - \frac{3}{2}x^2 + O(x^4))}{x^2} = \lim\limits_{x\to 0} \frac{bx^2+a+3 - \frac{9}{2}x^2 + O(x^4)}{x^2}.$$ Therefore you need to put $a = -3$ and $b = \frac{9}{2}$.