If $\alpha$ is positive root of the equation, p(x) = x2 - x - 2 = 0, then
$$\mathop {\lim }\limits_{x \to {\alpha ^ + }} {{\sqrt {1 - \cos \left( {p\left( x \right)} \right)} } \over {x + \alpha - 4}}$$ is equal to :
Solution
${x^2} - x - 2 = 0$<br><br>roots are 2 & $-$1 $\Rightarrow$ $\alpha$ = 2 (given $\alpha$ is positive)<br><br>Now $$ \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {1 - \cos ({x^2} - x - 2)} } \over {(x - 2)}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt {2{{\sin }^2}{{({x^2} - x - 2)} \over 2}} } \over {(x - 2)}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}$$<br><br>$= {3 \over {\sqrt 2 }}$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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