ISI2014-DCG-17

Question

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

• A. $0$
• B. $1/2$
• C. $1$
• D. non-existent

Answer 1

As  $x \to 2^-$ , $\dfrac{1}{x-2} \to -\infty$( so $e^{\frac{1}{x-2}} \to 0 $) , so $\lim_{x\to 2^-} \dfrac{1}{1+e^{\frac{1}{x-2}}} = \dfrac{1}{1+0} = 1$

As  $x \to 2^+$ , $\dfrac{1}{x-2} \to \infty$ ( so $e^{\frac{1}{x-2} }\to \infty $) , so $\lim_{x\to 2^+} \dfrac{1}{1+e^{\frac{1}{x-2}}} = 0 $

Left Hand limit $\neq$ Right hand limit, so limit DNE , option D is correct one.

Answer 2

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.

Comments

Right hand limit is zero but not the left hand limit which is 1 . Hence D should be answer.

Answer 3

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

Comments

Are you sure that as $x \to 2$, $\dfrac{1}{x-2} \to \infty$?