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If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c$, then evaluate $$\lim_{x\rightarrow \alpha}(1+ax^2+bx+c)^{\frac{1}{(x-\alpha)}}$$

Here's what I have tried:

$$\begin{eqnarray}\lim_{x\rightarrow \alpha}(1+ax^2+bx+c)^{\dfrac{1}{(x-\alpha)}} &=& \lim_{x\rightarrow \alpha}\exp\left(\frac{\ln(1+ax^2+bx+c)}{(x-\alpha)}\right) \\ &=&\exp\left(\lim_{x\rightarrow\alpha}\frac{\ln(1+ax^2+bx+c)}{(x-\alpha)}\right) \\ &=& \exp\left(\lim_{x\rightarrow\alpha}\frac{2ax+b}{1+ax^2+bx+c}\right)\\ &=&\exp\left(\frac{2a\alpha + b}{1+0}\right) \\ &=&\exp(2a\alpha+b)\end{eqnarray}$$

However, the solution to this problem is $e^{a(\alpha-\beta)}$. Is there something wrong in my approach?

Cameron Williams
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Vibhav
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2 Answers2

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Both are correct. You can convert into $answer's$ form by using $\alpha+\beta=-b/a$ using Vieta's theorem.(sum of roots)

evil999man
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Awesome's answer shows that your is right. Here is what you do to obtain it directly.

Write $$ax^2+bx+c=a(x-\alpha)(x-\beta)$$

Guy
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