Question

alpha, beta are the roots of the X square + bx + c = 0 then limit X tends to Alpha 1 + x square + bx + c 1 by x minus Alpha isif alpha, beta are the roots of x square + bx + c then Limit extends to Alpha 1 + x square + bx + c ^ 1 minus one by x minus Alpha is ​

Answer 1

Answer:

Given that

$\alpha \: is \: root \: of \: {x}^{2} + bx + c \\ so \\ { \alpha }^{2} + b \alpha + c = 0$

We have to find

$\lim_{x \to \alpha }(1 + {x}^{2} + bx + c ) {}^{\frac{1}{x - \alpha } } \: \: \\ \\ take \: log \\ \\ \lim_{x \to \alpha } \frac{ log(1 + {x}^{2} + bx + c ) }{x - \alpha } \\ \\ \: \: use \: \: l \: hopital \: (differentiate) \\ \\ \lim_{x \to \alpha } \frac{ \frac{2x + b}{1 + {x}^{2} + bx + c } }{1} \\ \\ = \frac{2 \alpha + b }{1 + { \alpha }^{2 } + b \alpha + c} \\ \\ = {2 \alpha + b \over1 + 0}\\\\=\huge\red{2 \alpha + b}$