# The behaviour of e^x near infinity and -infinity

1. Jun 1, 2008

### laura_a

1. The problem statement, all variables and given/known data

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?

2. Jun 1, 2008

### 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 $x \neq 0$).

Last edited: Jun 1, 2008
3. Jun 1, 2008

### 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?

4. Jun 1, 2008

### HallsofIvy

Staff Emeritus
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)?