F(x) = cases (e^ x^2 - 1) ( (1 + 2x)) ^6(2 x ) , & x 0 k, & x = 0 cases The value of…

MATHEMATICS · Single correct · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

f(x)={(ex2−1)sin⁡(ln⁡(1+2x))tan⁡6(2x),x≠0k,x=0f(x) = \begin{cases} \frac{(e^{x^2} - 1) \sin(\ln(1 + 2x))}{\tan^6(2\sqrt{x})}, & x \neq 0 \\ k, & x = 0 \end{cases} The value of kk for which f(x)f(x) is continuous at x=0x = 0, is (a) 124\frac{1}{2^4} (b) 125\frac{1}{2^5} (c) 123\frac{1}{2^3} (d) 126\frac{1}{2^6}

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