If $\alpha, \beta$ are the roots of the equation $lx^2+mx+n=0$ then $\lim\limits_{x \rightarrow \alpha}\left(l x^2+m x+n+1\right)^{\frac{1}{x-\alpha}}$ is 

$e^{l(\alpha-\beta)}$

$\alpha, \beta$ are roots of $lx^2+m x+n=0$, then the given equation can be written as

$lx^2+m x+n=0=l(x-\alpha)(x-\beta)$

Hence $\lim\limits_{X \rightarrow \alpha}\left(l x^2+m x+n+1\right)^{\frac{1}{x-\alpha}}$

$=e^{~\lim\limits_{x \rightarrow \alpha} \frac{l x^2+m x+n}{x-\alpha}}=e^{l^{(\alpha-\beta)}}$

Hence (4) is the correct answer.