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Limits & Continuityhard
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Determine the value of the constant kkk such that the function f(x)={ekx−1xx<03x+k2x≥0f(x) = \begin{cases} \frac{e^{kx}-1}{x} & x < 0 \\ 3x + k^2 & x \ge 0 \end{cases}f(x)={xekx−1​3x+k2​x<0x≥0​ is continuous at x=0x = 0x=0.