If \(m\) and \(n\) are the roots of the equation \(x^2 - px + q = 0\), then what is \(\displaystyle\lim_{x\to m} \frac{e^{x^2-px+q}-1}{(x-m)(x-n)}\) equal to?
\(1\)
Since \(m\) and \(n\) are roots of \(x^2 - px + q = 0\), we have \(x^2 - px + q = (x-m)(x-n)\). So the expression becomes \(\dfrac{e^{(x-m)(x-n)}-1}{(x-m)(x-n)}\). Putting \(t = (x-m)(x-n)\), as \(x \to m\), \(t \to 0\), and the limit reduces to the standard limit \(\displaystyle\lim_{t\to 0}\frac{e^t-1}{t} = 1\).
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