Let f(x)={x2+ax−2a−4x−2,x<24x−a,x≥2f(x)=\begin{cases} \dfrac{x^{2}+ax-2a-4}{x-2}, & x<2\\ 4x-a, & x\ge 2 \end{cases}…

Mathematics

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Let f(x)={x2+ax−2a−4x−2,x<24x−a,x≥2f(x)=\begin{cases} \dfrac{x^{2}+ax-2a-4}{x-2}, & x<2\\ 4x-a, & x\ge 2 \end{cases} Determine the value of aa for which lim⁡x→2f(x)\displaystyle\lim_{x\to 2}f(x) exists. a) a=4a=4 b) a=2a=2 c) a=3a=3 d) a=−2a=-2

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