Let x=2 be a root of the equation x^2+px+q=0 and f(x) = \left\{ {\begin{matrix} {{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x
e 2p} \ {0,} & {x = 2p} \ \end{matrix} } ight. Then \mathop {\lim }\limits_{x o 2{p^ + }} [f(x)], where \left[ . ight] denotes greates

Question

Let x=2 be a root of the equation x^2+px+q=0 and f(x) = \left\{ {\begin{matrix} {{{1 - \cos ({x^2} - 4px + {q^2} + 8q + 16)} \over {{{(x - 2p)}^4}}},} & {x 
e 2p} \ {0,} & {x = 2p} \ \end{matrix} } ight. Then \mathop {\lim }\limits_{x 	o 2{p^ + }} [f(x)], where \left[ . ight] denotes greatest integer function, is

Accepted Answer

Answer: C. 0 \lim \limits_{x \rightarrow 2^{+}}\left(\frac{1-\cos \left(x^{2}-4 p x+q^{2}+8 q+16\right)}{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}\right)\left(\frac{\left(x^{2}-4 p x+q^{2}+8 q+16\right)^{2}}{(x-2 p)^{2}}\right) \lim \limits_{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2} Using L'Hospital's \lim _{x \rightarrow 2 p^{+}}[f(x)]=0

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