if $\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$ then $6(\alpha+\beta)=?$ Let $\alpha, \beta \in \mathbb{R}$ be such that $$\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x} = 1$$ then what is the value of $6(\alpha+\beta)$
 A: Applying l'Hospital:
$$\lim_{x\to0}\frac{x^2\sin\beta x}{\alpha x-\sin x}=\lim_{x\to0}\frac{2x\sin\beta x+\beta x^2\cos\beta x}{\alpha-\cos x}$$
Since the limit in the numerator is zero, it must be the same in the denominator, so $\;\alpha=1\;$ ...Can you take it from here?
A: You can rewrite your function as
$$
\frac{\beta x^3}{\alpha x-\sin x}\frac{\sin(\beta x)}{\beta x}
$$
If $\alpha\ne 1$, this also becomes
$$
\frac{\beta x^2}{\alpha-\dfrac{\sin x}{x}}\frac{\sin(\beta x)}{\beta x}
$$
Note that the second factor has limit $1$ and the first factor has limit $0$ (the numerator has limit $0$; the denominator has limit $\alpha-1\ne0$).
Thus if the limit has to be $1$, we need $\alpha=1$.
Now it should be well known that
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=\frac{1}{6}
$$
(apply l'Hôpital or a Taylor expansion), so we have
$$
\lim_{x\to0}
  \frac{\beta}{\dfrac{x-\sin x}{x^3}}
  \frac{\sin(\beta x)}{\beta x}
=6\beta
$$
Now it's just an easy computation.
A: $$
\begin{aligned}
\lim _{x\to 0}\left(\frac{x^2\sin \left(\beta \:x\right)}{\alpha \:x-\sin \:x}\right)
& = \lim _{x\to 0}\left(\frac{x^2\left(\beta x+o\left(x\right)\right)}{\alpha \:x-\left(x-\frac{x^3}{3!}+o(x^3)\right)}\right)
\\& = \beta \:\cdot \lim _{x\to 0}\left(\frac{6x^2}{6α+x^2-6}\right) (\text{so } \color{red}{\beta \ne 0})
\\& = \beta \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = 6\beta   \text{ (with } \color{red}{\alpha = 1})
\\& = \color{red}{\frac{1}{6}} \:\cdot\lim _{x\to \:0}\left(\frac{6x^2}{6+x^2-6}\right) = \color{blue}{1}   \text{ (with } \color{red}{\beta = \frac{1}{6}})
\end{aligned}
$$
So, with
$$\alpha = 1, \beta = \frac{1}{6}$$
$$6(\alpha + \beta)=6\left(1+\frac{1}{6}\right) = \color{gold}{7}$$
A: HINT:
As $x\to0$, $\sin(\beta x)=\beta x+O(x^3)$ and $\sin(x)=x-\frac16 x^3+O(x^5)$
A: Apply L'Hospital's rule repeatedly. Doing so just once shows that $\alpha$ must be $1$. Then doing so two more times shows that $6 \beta$ is the limit, i.e., $6 \beta = 1$, or $\beta  = \tfrac{1}{6}$. Thus, $6(\alpha+\beta) = 6(1 + \tfrac{1}{6}) = 7$.
A: I tried to not use L'Hospitale to make the result more explicative: 
First you should note that $\beta$ is not zero, because the resulting limit is not one.
Second:
$$\lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x}=1 \iff \lim_{x\to 0} \frac{x^2\sin(\beta x)}{\alpha x-\sin x}-1=0$$
$$\iff \lim_{x\to 0} \frac{-x α + \sin(x) + x^2 \sin(x β)}{x α - \sin(x)}=0$$
$$\iff \lim_{x\to 0} \frac{-α + \frac{\sin(x)}{x} + x \sin(x β)}{α - \frac{\sin(x)}{x}}=0$$
$$\iff \lim_{x\to 0} \frac{-α + \frac{\sin(x)}{x} + x^2β \frac{\sin(x β)}{\beta x}}{α - \frac{\sin(x)}{x}}=0$$
if $\alpha$ is something different from 1 then the limit would not be $0$.
I think that the rest you should complete it.
A: Since the given limit $$\lim_{x \to 0}\frac{x^{2}\sin \beta x}{\alpha x - \sin x} = 1$$ is non-zero, it follows that both the numerator and denominator of the expression under limit operation are non-zero as $x \to 0$. Hence $\beta \neq 0$. Now taking reciprocals we see that $$\lim_{x \to 0}\frac{\alpha x - \sin x}{x^{2}\sin \beta x} = 1$$ or $$\lim_{x \to 0}\frac{\alpha x - \sin x}{\beta x^{3}}\cdot\frac{\beta x}{\sin \beta x} = 1$$ or $$\lim_{x \to 0}\frac{\alpha x - \sin x}{x^{3}} = \beta\tag{1}$$ We next use the well known limit $$\lim_{x \to 0}\frac{x - \sin x}{x^{3}} = \frac{1}{6}\tag{2}$$ (easily proved by Taylor's series expansion or L'Hospital's Rule). Subtracting $(2)$ from $(1)$ we get $$\lim_{x \to 0}\frac{\alpha - 1}{x^{2}} = \beta - \frac{1}{6}\tag{3}$$ and then $$\alpha - 1 = \lim_{x \to 0}\frac{\alpha - 1}{x^{2}}\cdot x^{2} = \left(\beta - \frac{1}{6}\right)\cdot 0 = 0$$ so we can see that $\alpha = 1$ and from $(3)$ we now get $\beta = 1/6$. Thus $6(\alpha + \beta) = 7$.

Most answers here try to deduce $\alpha = 1$ by using contradiction. It is much easier to use algebra of limits directly to reach the conclusion $\alpha = 1$.
