Notifications
Clear all
1
05/01/2022 2:08 pm
Topic starter
If α, β are distinct roots of x2 + bx + c = 0 then find
\(\lim\limits_{x \to \beta}\frac{e^{2(x^2 +bx+c)}-1-2(x^2 +bx+c)}{(x-\beta)^2}\)
This topic was modified 3 years ago by admin
1 Answer
1
05/01/2022 2:16 pm
\(\lim\limits_{x \to \beta}\frac{e^{2(x^2 +bx+c)}-1-2(x^2 +bx+c)}{(x-\beta)^2}\)
= \(\lim\limits_{x \to \beta}\frac{e^{2(x-\alpha)(x - \beta)}-1-2(x-\alpha)(x-\beta)}{(x-\beta)^2(x-\alpha)^2}\) x (x - α)2
Let (x - α)(x - β) = y
then \(\lim\limits_{y \to 0}\frac{e^{2y}-1-2y}{y^2}\) x (β - α)2
Using L hospital rule,
= (β - α)2\(\lim\limits_{y \to 0}\frac{2e^{2y}-2}{2y}\)
= (β - α)2\(\lim\limits_{y \to 0}\frac{e^{2y}-1}{2y}\) x 2
This implies, 2(β - α)2 = 2(b2 - 4c)