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

Thread Starter

magnet18

Joined Dec 22, 2010
1,227
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)^∞ = ∞
 

Thread Starter

magnet18

Joined Dec 22, 2010
1,227
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 :p
 

WBahn

Joined Mar 31, 2012
29,978
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)^∞ = ∞
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.
 
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