Show that 1+kx is the local linearization of (1+x)^k near x = 0

Local linearization L(x) of a function f(x) near x = a is given by f(a) + f'(a)(x - a)
f(x) = (1 + x)^k
f(0) = (1 + 0)^k = 1^k = 1
f'(x) = k(1 + x)^(k - 1)
f'(0) = k(1 + 0)^(k - 1) = k(1)^(k - 1) = k(1) = k

Therefore, the local linearization L(x) of (1 + x)^k near x = 0 is given by f(0) + f'(0)(x - 0) = 1 + k(x) = 1 + kx

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