If $\alpha ,\beta$ are two distinct real roots of the equation $ax^3+x-1-a=0,(a\neq -1,0)$.none of which is equal to unity ,then the value of $\displaystyle \lim_{x\rightarrow (1/a)}\dfrac{(1+a)x^3-x^2-a}{(e^{1-ax}-1)(x-1)}$ is $\dfrac{aL(k\alpha -\beta}{\alpha}$ the the value of $KL$ is : ?

Factoring the numerator gives ((x^3 - 1) + a(x^3 - 1)), which factors to (x - 1)(x^2 + x + 1)(ax + 1). Evaluating the limit as x approaches 1/a involves substituting 1/a into the factored numerator to get (1/a - 1)(1/a^2 + 1/a + 1)(1 + 1). Using the property of polynomial roots, since alpha is a root of ax^3 + x - 1 - a = 0, the expression (1/a^2 + 1/a + 1) simplifies to (beta + 1) divided by alpha. The limit evaluates to (1 - a)(beta + 1)(a + 1) all divided by a^2 times (e^(1 - 1) - 1), which simplifies to (1 - a^2)(beta + 1) divided by a. This yields a value of negative L times (beta + 1) divided by alpha. Equating this to the given form (aL(k times alpha minus beta)) divided by alpha and setting a equal to 1 shows that k equals 1.

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