If f (x) = (x + 1)cot x be continuous at x = 0, then f (0) is equal to

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Suppose f\left( x ight) is continuous at x = 0 . Given: f\left( x ight) = \left( x + 1 ight)^{	ext{ cot } x} \log f\left( x ight) = \left( \cot x ight) \left( \log \left( x + 1 ight) ight) \left[ 	ext{ Taking log on both sides } ight] \Rightarrow \lim_{x 	o 0} \log f\left( x ight) = \lim_{x 	o 0} \left( \cot x ight) \left( \log \left( x + 1 ight) ight) \Rightarrow \lim_{x 	o 0} \log f\left( x ight) = \lim_{x 	o 0} \left( \frac{\log \left( x + 1 ight)}{	an x} ight) \Rightarrow \lim_{x 	o 0} \log f\left( x ight) = \lim_{x 	o 0} \frac{\left( \frac{\log \left( x + 1 ight)}{x} ight)}{\left( 	an x/x ight)} \Rightarrow \lim_{x 	o 0} \log f\left( x ight) = \frac{\lim_{x 	o 0} \left( \frac{\log \left( x + 1 ight)}{x} ight)}{\lim_{x 	o 0} \left( 	an x/x ight)} \Rightarrow \log \left( \lim_{x 	o 0} f\left( x ight) ight) = \frac{\lim_{x 	o 0} \left( \frac{\log \left( x + 1 ight)}{x} ight)}{\lim_{x 	o 0} \left( 	an x/x ight)} \left[ \because f\left( x ight)	ext{ is continuous at } x = 0 ight] \Rightarrow \log \left( \lim_{x 	o 0} f\left( x ight) ight) = 1 \Rightarrow \lim_{x 	o 0} f\left( x ight) = e \Rightarrow f\left( 0 ight) = e \left[ \because f\left( x ight) 	ext{ is continuous at } x = 0 ight]

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