Question

I just need a little clue..... continuity/limits
Graphing differs from my by-hand conclusion.
Should be not hard....

Answer 1

\(\Large\color{blue}{ f(x)=\frac{1}{1+e^{1/x}}\LARGE\color{white}{ \rm │ }}\)

Answer 2

Continuity

When I am graphing, I am getting that lim x→0 Does Not Exist
But, I am showing though, that

x → 0\(\scriptsize\color{slate}{^-}\)
1/x → 0
e \(\scriptsize\color{slate}{^{1/x}}\) → 1
1+ e \(\scriptsize\color{slate}{^{1/x}}\) → 2
So it approaches 1/2 from the left.

Answer 3

What do you need help with?

Answer 4

I am showing, hold on

Answer 5

This is how a graph for it looks, by the way

Answer 6

x→0 \(\scriptsize\color{slate}{^{+}}\)
(1/x) → ∞
e^(1/x) → ∞
denominator → ∞
f(x) → 0


x→0 \(\scriptsize\color{slate}{^{-}}\)
(1/x) → 0
e^(1/x) → 1
denominator → 2
f(x) → 1/2


however desmos gives me
https://www.desmos.com/calculator/mqu6gcfm52

Answer 7

Was I right or wrong, my friend?

Answer 8

wrong, it doesn't approach the same value from left and right side as x approaches zero.

Answer 9

I am just asking that should it be approaching 1/2 from the left side, not 1 ?

Answer 10

e^(1/x) can't → 0

Answer 11

I believe you have to make the approaching of the left 1 by the side, as the calculation basically makes it go a bit higher, like one point, but drops it (Sorry for my bad English)

Answer 12

Make it approach ?

Answer 13

Eh, I am not a very good with speaking math, just doing it

Answer 14

\[x\to 0\implies \frac{1}{x}\to 0\]??

Answer 15

FROM THE LEFT SIDE, YES

Answer 16

I am just asking that

x → 0 \(\scriptsize\color{slate}{ ^{-} }\)

1/x → 0

e\(\normalsize\color{slate}{ ^{1/x} }\) → 1

denominator → 2

f(x) → 1/2

Answer 17

The graph indicates that it approaches 1 from the left though, but e^(1/x) can't be approaching zero, can it ?

Answer 18

e^0 = 1 ?

Answer 19

Re-posting the function, \(\LARGE\color{black}{ f(x)=\frac{1}{1+e^{1/x}} }\)

Answer 20

depends on the direction
\[x\to 0+\implies \frac{1}{x}\to +\infty \implies e^{\frac{1}{x}}\to \infty\implies f\to 0\]

Answer 21

\[x\to 0^-\implies \frac{1}{x}\to -\infty \implies e^{\frac{1}{x}}\to 0\implies f\to 1\]

Answer 22

x → 0 ^-
1/x → 1/-00.1 → -∞
e^1/x → e^(-1000) → e^1/1000 → e^0 → 1.