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ISI2014DCG17
+2
votes
30
views
$\underset{x \to 2}{\lim} \dfrac{1}{1+e^{\frac{1}{x2}}}$ is
$0$
$1/2$
$1$
nonexistent
isi2014dcg
calculus
limits
asked
Sep 23, 2019
in
Calculus
by
Arjun
Veteran
(
431k
points)
recategorized
Nov 9, 2019
by
Lakshman Patel RJIT

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1
Answer
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Put x=2 in the power of e, we get e^(1/(22)) which is e^infinity which is infinity.. Now 1/(1+infinity) = 1/ infinity which is equal to zero. Hence answer is zero.
answered
Sep 26, 2019
by
kp1
(
143
points)
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ISI2014DCG2
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ISI2014DCG33
Let $f(x)$ be a continuous function from $[0,1]$ to $[0,1]$ satisfying the following properties. $f(0)=0$, $f(1)=1$, and $f(x_1)<f(x_2)$ for $x_1 < x_2$ with $0 < x_1, \: x_2<1$. Then the number of such functions is $0$ $1$ $2$ $\infty$
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isi2014dcg
calculus
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