# Let $\dpi{100} \alpha \: and\: \beta$ be the distinct roots of $\dpi{100} ax^{2}+bx +c =0,$Then $\dpi{100} \lim_{x\rightarrow \alpha }\frac{1-\cos \left ( ax^{2} +bx+c\right )}{\left ( x-\alpha \right )^{2}}$ is equal to Option 1) $0$ Option 2) $\frac{a^{2}}{2}\left ( \alpha -\beta \right )^{2}$ Option 3) $\frac{1}{2}\left ( \alpha -\beta \right )^{2}$ Option 4) $\frac{-a^{2}}{2}\left ( \alpha -\beta \right )^{2}$

H Himanshu

As we have learned

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow \alpha } \frac{1- \cos (ax^2 +bx+c)}{(x-\alpha )^2}$

$\lim_{x\rightarrow \alpha } \frac{1- \cos (a(x-\alpha )^2 (x-\beta ))}{(x-\alpha )^2}$

$= \lim_{x\rightarrow \alpha } \frac{2\sin ^2\frac{ (a(x-\alpha )^2 (x-\beta ))}{2} }{(x-\alpha )^2}$

$= \lim_{x\rightarrow \alpha }2\left ( \frac{\sin \frac{ (a(x-\alpha )^2 (x-\beta ))}{2} }{\frac{ (a(x-\alpha )^2 (x-\beta ))}{2}}\right )^2\times \left ( \frac{a^2}{4}(x-\beta )^2 \right )$

$2\times 1\times \frac{a^2}{4}(\alpha -\beta )^2 = \frac{a^2(\alpha -\beta )^2}{2}$

Option 1)

$0$

Option 2)

$\frac{a^{2}}{2}\left ( \alpha -\beta \right )^{2}$

Option 3)

$\frac{1}{2}\left ( \alpha -\beta \right )^{2}$

Option 4)

$\frac{-a^{2}}{2}\left ( \alpha -\beta \right )^{2}$

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