If-are-root-of-quadratic-equation-ax-2-bx-c-then-lim-x-1-cos-ax-2-bx-c-x-2-

Question Number 63920 by raj last updated on 11/Jul/19

If α, β are root of quadratic equation

{a x}^{2} + b x + c then

\lim_{x \to α} \frac{1 - \cos ({a x}^{2} + b x + c)}{{(x - α)}^{2}} = ?

Commented by Prithwish sen last updated on 11/Jul/19

= Lt {[\frac{\sin \frac{(x - α) (x - β)}{2}}{\frac{(x - α) (x - β)}{2}}]}^{2} . \frac{{(x - β)}^{2}}{2}

= \frac{{(α - β)}^{2}}{2}

please check .

Commented by kaivan.ahmadi last updated on 11/Jul/19

h o p

{l i m}_{x \to α} \frac{(2 a x + b) s i n ({a x}^{2} + b x + c)}{2 (x - α)} \overset{h o p}{=}

{l i m}_{x \to α} \frac{{(2 a x + b)}^{2} c o s ({a x}^{2} + b x + c)}{2} = \frac{{(2 a α + b)}^{2}}{2}

Commented by mathmax by abdo last updated on 11/Jul/19

l e t u (x) = {a x}^{2} + b x + c w e h a v e {l i m}_{x \to α} u (x) = 0 a n d

1 - c o s (u (x)) \sim \frac{u^{2} (x)}{2} (x \in V (α)) \Rightarrow

\frac{1 - c o s u (x)}{{(x - α)}^{2}} \sim \frac{u^{2} (x)}{2 {(x - α)}^{2}} = \frac{{(a (x - α) (x - β))}^{2}}{2 {(x - α)}^{2}} = \frac{a^{2} {(x - β)}^{2}}{2} \Rightarrow

{l i m}_{x \to α} \frac{1 - c o s u (x)}{{(x - α)}^{2}} = \frac{a^{2}}{2} {(α - β)}^{2} .

Commented by raj last updated on 11/Jul/19

t h a n k y o u

Commented by mathmax by abdo last updated on 12/Jul/19

y o u a r e w e l c o m e .