The value of `f(0)` so that the function `f(x)=(2-(256-7x)^(1/8))/((5x+32)^(1//5)-2),x!=0` is continuous everywhere, is given by

Options

A. (a) `-`1
B. (b) 1
C. (c) 26
D. (d) none of these

Explanation

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{2 - (256 - 7x)^{1/8}}{(5x + 32)^{1/5} - 2}, \quad x \neq 0 \]

is continuous everywhere, we need to ensure that the limit of \( f(x) \) as \( x \) approaches 0 equals \( f(0) \).

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