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  • The value of f(0), so that the function <img alt="f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}}(x \neq 0)" src="https://entrancecorner.oncodecogs.com/gif.la

The value of f(0), so that the function f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}}(x \neq 0) is continuous is

Option: 1

2/3


Option: 2

6


Option: 3

2


Option: 4

4


Answers (1)

best_answer

\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{3\left[3-(243+5 x)^{1 / 5}\right]}=\lim _{x \rightarrow 0} \frac{\frac{(27-2 x)^{1 / 3}-(27)^{1 / 3}}{(27-2 x)-27}(-2 x)}{-3\left[\frac{(243+5 x)^{1 / 5}-(243)^{1 / 5}}{243+5 x-243} \cdot 5 x\right]} \\=\frac{2}{15} \frac{\frac{1}{3}(27)^{-2 / 3}}{\frac{1}{5}(243)^{-4 / 5}}=\frac{2}{15} \cdot \frac{5}{3} \cdot \frac{1}{9} \cdot 81=2

Hence (c) is the correct answer.

 

Posted by

Gaurav

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