A) \[2/3\]
B) 6
C) 2
D) 4
Correct Answer: C
Solution :
Since \[f(x)\] is continuous at \[x=0,\] therefore \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{(27-2x)}^{1/3}}-3}{9-3\,{{(243+5x)}^{1/5}}}\,\], \[\left( \text{Form}\frac{\text{0}}{\text{0}} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{\frac{1}{3}\,{{(27-2x)}^{-2/3}}(-2)}{-\frac{3}{5}\,{{(243+5x)}^{-4/5}}(5)}=2.\]
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