<s>[INST] $\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)x^3-x^2-a}{(e^{1-\alpha x}-1)(x-1)}$ is $\frac{al(k\alpha-\beta)}{\alpha}$ If $\alpha,\beta$ are two distinct real roots of the equation $ax^3+x-1-a=0,(a\ne-1,0)$,none of which is equal to unity,then $\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)x^3-x^2-a}{(e^{1-\alpha x}-1)(x-1)}$ is $\frac{al(k\alpha-\beta)}{\alpha}$.Find $kl$.

Since $\alpha,\beta$ are the roots of the equation $ax^3+x-1-a=0$.So $a\alpha^3+\alpha-1-a=0$ and $a\beta^3+\beta-1-a=0$
$\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)x^3-x^2-a}{(e^{1-\alpha x}-1)(x-1)}$
$\lim_{x\to\frac{1}{\alpha}}\frac{(1+a)\frac{1}{\alpha^3}-\frac{1}{\alpha^2}-a}{(e^{1-\alpha x}-1)(x-1)}$
I am stuck here.The numerator is tending to zero and the denominator is tending to zero,i applied L Hospital rule,but could not solve it.
 [/INST] First of all, since we have that
$$ax^3+x-1-a=(x-1)(ax^2+ax+a+1)$$
and that $\alpha\not=1,\beta\not=1$, we can have that
$$a\alpha^2+a\alpha+a+1=a\beta^2+a\beta+a+1=0,$$
i.e.
$$(1+a)\left(\frac{1}{\alpha}\right)^2+a\cdot\frac{1}{\alpha}+a=(1+a)\left(\frac{1}{\beta}\right)^2+a\cdot\frac{1}{\beta}+a=0.$$
Since the numerator can be factored as
$$(1+a)x^3-x^2-a=(x-1)((1+a)x^2+ax+a)$$
we have
$$\lim_{x\to 1/\alpha}\frac{(x-1)((1+a)x^2+ax+a)}{(e^{1-\alpha x}-1)(x-1)}=\lim_{x\to 1/\alpha}\frac{(1+a)x^2+ax+a}{e^{1-\alpha x}-1}$$
Using l'Hôpital's rule once,
$$\begin{align}\lim_{x\to 1/\alpha}\frac{(1+a)x^2+ax+a}{e^{1-\alpha x}-1}&=\lim_{x\to 1/\alpha}\frac{2(1+a)x+a}{e^{1-\alpha x}(-\alpha)}\\&=\frac{2(1+a)(1/\alpha)+a}{e^{1-\alpha(1/\alpha)}(-\alpha)}\\&=\frac{-2-2a-a\alpha}{\alpha^2}\end{align}$$
So, noting that $\alpha+\beta=-a/a=-1\Rightarrow \beta=-\alpha-1$, we have
$$\frac{-2-2a-a\alpha}{\alpha^2}=\frac{al(k\alpha-\beta)}{\alpha}=\frac{al(k\alpha^2+\alpha^2+\alpha)}{\alpha^2}$$
$$\Rightarrow -2-2a-a\alpha=al(k\alpha^2+\alpha^2+\alpha)$$
Using $\alpha^2=-\alpha-1-\frac 1a$ and simplifying it
$$\alpha(2a+a^2-a^2l-al-alk)=a^2+a-lka^2-lka$$
$$\Rightarrow\alpha(2+a-al-l-lk)=a+1-lka-lk$$
Now using that if $p\alpha=q$, then $p^2(a\alpha^2+a\alpha+a+1)=0\Rightarrow aq^2+apq+p^2(a+1)=0$,
$$a(a+1-lka-lk)^2+a(a+1-lka-lk)(2+a-al-l-lk)+(a+1)(2+a-al-l-lk)^2=0$$
Expanding the LHS,
$$(k^2l^2+kl^2-3kl+l^2-3l+3)a^3+(3k^2l^2+4kl^2-10kl+3l^2-10l+10)a^2+(3k^2l^2+5kl^2-11kl+3l^2-11l+11)a+k^2l^2+2kl^2-4kl+l^2-4l+4=0$$
$$\Rightarrow k^2l^2+kl^2-3kl+l^2-3l+3=3k^2l^2+4kl^2-10kl+3l^2-10l+10=0$$
Now, noting that
$$3(k^2l^2+kl^2-3kl+l^2-3l+3)-(3k^2l^2+4kl^2-10kl+3l^2-10l+10)=0\Rightarrow (l-1)(kl-1)=0\Rightarrow l=1\quad\text{or}\quad kl=1$$
In either case, we have $k=l=1$, and this is sufficient.
Hence, the answer is $\color{red}{kl=1}$.
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