If $\alpha , \beta$ be the roots of $ax^2+bx+c=0$. Find $\lim_{x \to \alpha}[1+ax^2+bx+c]^\frac{1}{x-\alpha}$
Here $\alpha +\beta=-\frac{b}{a}$ and $\alpha \beta=\frac{c}{a}$. How can I proceed?
If $\alpha , \beta$ be the roots of $ax^2+bx+c=0$. Find $\lim_{x \to \alpha}[1+ax^2+bx+c]^\frac{1}{x-\alpha}$
Here $\alpha +\beta=-\frac{b}{a}$ and $\alpha \beta=\frac{c}{a}$. How can I proceed?
Since $\alpha$ and $\beta$ are roots, we have $ax^2+bx+c=a(x-\alpha)(x-\beta)$ So$\begin{align} \lim_{x \to \alpha}[1 + ax^2 + bx + c]^{\frac{1}{x-\alpha}} &= \lim_{x \to \alpha}[1 + a(x-\alpha)(x-\beta)]^{\frac{1}{x-\alpha}} \\ &= \lim_{y \to 0}[1 + y]^{{\frac{1}{y}}{\lim_{x \to \alpha}a(x-\beta)}}\\ &= e^{a(\alpha-\beta)} \end{align}$
$\lim_{x \rightarrow \alpha}{(1+ax^2+bx+c)^\dfrac{1}{x-\alpha}}=\lim_{x \rightarrow \alpha}{e^{\dfrac{ln(1+ax^2+bx+c)}{x-\alpha}}}=\lim_{x \rightarrow \alpha}{e^{\dfrac{2ax+b}{1+ax^2+bx+c}}}=e^{2a\alpha+b}$