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In order that the function $f(x)=(x+1)^{\large\cot x}$ is continuous at $x=0$. f(0) must be defined as

$(a)\;f(0)=0\qquad(b)\;2\qquad(c)\;f(0)=\large\frac{1}{e}$$\qquad(d)\;None\;of\;these$

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$l=\lim\limits_{x\to 0}(x+1)^{\large\cot x}$
$\Rightarrow \log l=\lim\limits_{x\to 0}\cot x.\log(1+x)$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\log(1+x)}{\tan x}\qquad\big(\large\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{\Large\frac{1}{1+x}}{\sec^2x}$$=1$
$l=e^1=e$
For continuity value=limit
$\therefore f(0)=e$
Hence (d) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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