End behavior of e^(-x) + e^[x + e^(-x)]

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Dorkus73

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In class we covered behavior of infinity of e equations. I came up with the following equation e^(-x) + e^[x + e^(-x)]. You can simplify it to e^[ e^(-x) ] so if x →+∞ f(x)→ 1 because e^(-x) approaches 0 and e^0 = 1, but if you try a large number for x like 100 you get around 2.69 * 10 ^ 43 (I could't go much further as the calculator just said infinity/overflow). I assume that the graph is going to grow and end up at 1 when you "get to infinity"(in quotation marks because infinity isn't really a number that you can reach from my understanding) but my teacher said that f(x) is going to start decreasing at one point and is going to approach 1. Is it true that the graph/f(x) at one point stops growing and starts approaching 1 and if yes, can you find out when the graph does start to approach 1. (Sorry if my english is bad and if I can't describe the problem well enough, english isn't my first language)

Bild_2023-11-13_190902177.png
If this question is unfit for beginning algebra then I can post it in Intermediate algebra, I'm unfarmilliar with the classifications.
 
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I just realised my mistake, I wrote down the wrong equation and there is no problem anymore. Sorry for wasting your time. How can I delete the post?
 
Dorkus73 said:
I came up with the following equation e^(-x) + e^[x + e^(-x)]. You can simplify it to e^[ e^(-x) ] so if x →+∞ f(x)→ 1 because e^(-x) approaches 0 and e^0 = 1, but if you try a large number for x like 100 you get around 2.69 * 10 ^ 43 (I could't go much further as the calculator just said infinity/overflow). I assume that the graph is going to grow and end up at 1 when you "get to infinity"(in quotation marks because infinity isn't really a number that you can reach from my understanding) but my teacher said that f(x) is going to start decreasing at one point and is going to approach 1.
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I'm a little confused, because you say you think the limit is 1, but your "my" graph approaches infinity.

Since your simplification is wrong (it isn't true that a^m + a^n = a^(m+n)), we have to go by the original form, e^(-x) + e^[x + e^(-x)]. The first term goes to 0, while the second approaches e^x, which goes to infinity. So the sum will go to infinity, as in your graph.

Of course, we can actually graph it (e.g. in Desmos) and find out:

1699904605769.png​

Now, if we consider e^[ e^(-x) ] alone, then you are right that it should approach 1. It should also approach infinity to the left. Here is the graph:
1699905038445.png​

But it doesn't have to rise and fall.

Dorkus73 said:
If this question is unfit for beginning algebra then I can post it in Intermediate algebra, I'm unfamiliar with the classifications.
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It really belongs under calculus, where limits are properly studied; exponential functions probably belong under intermediate algebra or higher, but limits are not part of algebra. (But it doesn't really matter where you put anything; we don't get so many submissions that anyone looks only under their favorite category.)

The important thing to learn for the moment is how to simplify exponential expressions. They don't always simplify as far as you wish they did! And limits are not always obvious. (That's why it's a separate field!)
 
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