Functions Question 191
Question: The value of $ f(0) $ , so that the function $ f(x)=\frac{{{(27-2x)}^{1/3}}-3}{9-3{{(243+5x)}^{1/5}}},,(x\ne 0) $ is continuous, is given by
Options:
A) $ 2/3 $
B) 6
C) 2
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
Since $ f(x) $ is continuous at $ x=0, $ therefore $ f(0)=\underset{x\to 0}{\mathop{\lim }}f(x)=\underset{x\to 0}{\mathop{\lim }}\frac{{{(27-2x)}^{1/3}}-3}{9-3,{{(243+5x)}^{1/5}}}, $ , $ ( Form\frac{0}{0} ) $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{\frac{1}{3},{{(27-2x)}^{-2/3}}(-2)}{-\frac{3}{5},{{(243+5x)}^{-4/5}}(5)}=2. $
BETA
