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  • The value of f(0) so that the function  <img alt="\mathrm{f(x)=\frac{2-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-2}(x \neq 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmath

The value of f(0) so that the function  \mathrm{f(x)=\frac{2-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-2}(x \neq 0)}    is continuous everywhere; is given by

Option: 1

- 1


Option: 2

1


Option: 3

2^{6}


Option: 4

none of these


Answers (1)

best_answer

For f(x) to be continuous everywhere, we must have,
\mathrm{ f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-2} \quad\left(\text { Form } \frac{0}{0}\right) }

\mathrm{ =\lim _{x \rightarrow 0} \frac{\frac{7}{8}(256-7 x)^{-\frac{7}{8}}}{(5 x+32)^{-4 / 5}}=\frac{7}{8} \times \frac{2^{-7}}{2^{-4}}=\frac{7}{64} . }

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Gaurav

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