A) \[-1\]
B) \[1\]
C) \[{{2}^{6}}\]
D) None of these
Correct Answer: D
Solution :
For\[f(x)\]to be continuous everywhere, we must have \[f(0)=\underset{x\to 0}{\mathop{\lim }}\,f(x)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2-{{(256-7x)}^{1/8}}}{{{(5x+32)}^{1/5}}-2}\] \[=\lim \frac{\frac{7}{8}{{(256-7x)}^{-7/8}}}{{{(5x+32)}^{-4/5}}}\] \[=\frac{7}{8}\times \frac{{{2}^{-7}}}{{{2}^{-4}}}=\frac{7}{64}\]You need to login to perform this action.
You will be redirected in
3 sec