A) 1
B) e
C) \[\frac{1}{e}\]
D) \[{{e}^{2}}\]
Correct Answer: B
Solution :
[b] For a function to be continuous at a point the limit should exist and should be equal to the value of the function at that point. Here point is x = 0 and \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,{{(x+1)}^{\cot x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{(x+1)}^{\cot x}}=\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{\frac{1}{x}.x\cot x}}\] \[=\underset{x\to 0}{\mathop{\lim }}\,{{(x+1)}^{\frac{1}{x}\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\tan \,x}}}={{e}^{1}}=e\] Since limiting value of \[f(x)=e\], when \[x\to 0,f(0)\] should also be equal to e.
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