Question

Find the following limits for f(x)=1/(1+e^(1/x)).
a) limit as x approaches infinity of f(x)
b) limit as x approaches 1 of f(x)
c) limit as x approaches 0 from the right of f(x)
d) limit as x approaches 0 from the left of f(x)
e) limit as x approaches 0 of f(x)

Answer 1

a)\[\lim_{x \rightarrow \infty} \frac{ 1 }{ 1+e ^{\frac{ 1 }{ x }} }\]If x if off to infinity, 1/x becomes 0. So e^(1/x) becomes e^0 = 1. The limit is 1/(1+1) = 1/2.

Answer 2

Thank you very much. Unfortunately the part of this problem I'm really having trouble with is understanding how to solve c and d, though from looking at a graph it's obvious what the answers are. Is there a method to solve those two, or can it only be estimated visually?

Answer 3

c.
It is all about the behaviour of 1/x. If x goes to 0 from the right, 1/x becomes i.e. 1/0.1 = 10, 1/0.01 = 100, 1/0.001 = 1000 and so on. In short, 1/x goes to infinity.
Therefore e^(1/x) goes to infinity as well, because the larger the exponent of e^..., the larger the outcome. Adding 1 in the denominator of f doesn't change much.
The denominator also gets larger and larger.
1/(a larger and larger number) goes zo 0, so that's the value of your limit.

Answer 4

d.
Approaching from the left to 0 means that 1/x becomes very a large negative number:
1/-0.1 = -10, 1/-0.01 = -100, 1/-0.001 = -1000 and so on.
Now how does the exponential function react to this?
Well, e^(-1000) = 1/e^1000, so 1/(getting larger and larger) >> goes to 0.
The limit is 1/(1 + 0) = 1

Answer 5

That actually makes a lot of sense. Thank you for explaining it like that.

Answer 6

YW!