ISI2014-DCG-17

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$\underset{x \to 2}{\lim} \dfrac{1}{1+e^{\frac{1}{x-2}}}$  is

1. $0$
2. $1/2$
3. $1$
4. non-existent
in Calculus
recategorized

Put x=2 in the power of e, we get e^(1/(2-2)) which is e^infinity which is infinity.. Now 1/(1+infinity) = 1/ infinity which is equal to zero. Hence answer is zero.
1
Right hand limit is zero but not the left hand limit which is 1 . Hence D should be answer.
It is quite easy. When we put x=2 then e will became e^1/0 which will became infinity. Then 1/(1+∞) which is equal to Zero(0).

So Option C

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