# why doesn't lim x infinity (1+2/x)^x = infinity?

Discussion in 'Math' started by magnet18, Nov 3, 2012.

1. ### magnet18 Thread Starter Senior Member

Dec 22, 2010
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Why?
I found that the actual limit is e^2 using LHospitals rule, but why isn't it just infinity???
From what I see, lim x → infinity (1+2/x)^x = (1+(2/∞))^∞ = (1+0)^∞ = ∞

2. ### magnet18 Thread Starter Senior Member

Dec 22, 2010
1,232
124
I'm assuming it's some rule about x being an exponent in a limit, or something like that
possibly because you can't distribute the ^x to the 0, because 0^infinity is indeterminate?

I would just like to know at what point in solving this i need to decide that i need to use e and Ln rather than simplification and plugging in

exam tuesday

3. ### WBahn Moderator

Mar 31, 2012
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4,804
Why would (1+0)^∞ = ∞?

What's 1^10? 1^100? 1^100000000?

You don't know the 2/x is approaching zero "faster" than x is approaching infinity. If it is, then the answer will approach 1, if it isn't, then it will approach infinity, and if they two are "balanced" just right, it could converge on anything inbetween. In other words, it's indeterminate.

y = (1+2/x)^x

ln(y) = x * ln(1+2/x)

As x=>∞, this goes to ln(y) => ∞*0 which is indeterminate, so we can use L'Hospital's rule.

4. ### magnet18 Thread Starter Senior Member

Dec 22, 2010
1,232
124
OOOOOoooooooo...
missed that bit
*facepalm*
thanks