The behaviour of e^x near infinity and -infinity

[laura_a]

Homework Statement

I have done an integration and ended up with the result

[-c/2 * [e^(-2x)]] |^infinity_0 = 1
The solution is that c=2 so that means to me that e^(2x) must turn into minus 1 for it to equal 1... but I'm not sure.. I've got graphcalc so I've been staring at the graph and I figure that as x goes to infinity that e^x goes to 1... but not sure what to say when x goes to minus infinity?

 

[CompuChip]

You should remember the following properties of the exponential function:

• \lim_{x \to +\infty} e^x = \infty

• \lim_{x \to -\infty} e^x = 0

• \lim_{x \to 0} e^x = 1 (actually, the exponential function is continuous in 0, so one could also just say e^0 = 1, which is logical since x^0 = 1 for any x \neq 0).

 

[spastic]

e^x goes to 1 as x goes to 0.
e^x goes to 0 as x goes to negative infinity
e^x goes to infinity as x goes to infinity (no limit)

Is that what you're after?

 

[HallsofIvy]

Homework Statement

I have done an integration and ended up with the result

[-c/2 * [e^(-2x)]] |^infinity_0 = 1
The solution is that c=2 so that means to me that e^(2x) must turn into minus 1 for it to equal 1... but I'm not sure.. I've got graphcalc so I've been staring at the graph and I figure that as x goes to infinity that e^x goes to 1... but not sure what to say when x goes to minus infinity?

Then you need a new calculator! e^x does not go anywhere near 1 as x goes to infinity.
If you must use a calculator, what is e^1000000? What is e^(-100000)?