# Find $\lim_{x \to \alpha}[1+ax^2+bx+c]^\frac{1}{x-\alpha}$

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?

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The fact that $\beta$ is the name of the other root seems irrelevant, so it is odd it is mentioned explicitly in the question. –  TMM Dec 11 '12 at 13:13
@TMM:The answer of limit may contain $\beta$.You are right it is mentioned explicitly in the question. –  Argha Dec 11 '12 at 13:20
You could try taking a log and using L'Hospital's rule. –  nayrb Dec 11 '12 at 13:28

## 2 Answers

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}

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Thank you for your answer.I like it. –  Argha Dec 12 '12 at 14:42

$\lim_{x \rightarrow \alpha}{(1+ax^2+bx+c)^\dfrac{1}{x-\alpha}}=\lim_{x \rightarrow \alpha}{e^{\dfrac{ln(1+ax^2+bx+c)}{x-\alpha}}}=\lim_{x \rightarrow \alpha}{e^{\dfrac{2ax+b}{1+ax^2+bx+c}}}=e^{2a\alpha+b}$

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