What is the limit of (e^x + 2) ^1/x as x approaches to positive infinity?

Question

What is the limit of (e^x + 2) ^1/x as x approaches to positive infinity?

Answer 1

e^x for x very large is much larger than 2.

So limit of e^x+2 as x-> infinity = e^x.

So we're left with the limit of (e^x)^(1/x) as x goes to infinity.

^(1/x) reverses ^x so we just end up with e.

Answer 2

all you have to do is plug in big numbers for x and compute.

if you have a graphic calculator, 230 is the biggest number than you can plug in for x. any number bigger than that will say overflow.

anyways.

the answer to your question is:

the limit of (e^x + 2) ^1/x as x approaches to positive infinity is e.

i tried with 3 different values of x:

X=100

(e^100 + 2) ^1/100 = 2.7182…..

x=150

(e^150 + 2) ^1/150 = 2.7182…..

x= 230.

(e^230 + 2) ^1/230 = 2.7182…..

all of the above values of x give the same answer.

hope that helped

Answer 3

Using Microsoft Excel

=(EXP(100)+2)^1/100 gives 2.68812 x 10^41

=(EXP(709)+2)^1/709 gives 1.1592 x 10^305

Any value above 709 gives a #NUM! error

However

=(EXP(100)+2)^(1/100) gives 2.718281828 = e

=(EXP(709)+2)^(1/709) also gives e

Answer 4

♥ y=(e^x + 2) ^1/x; z=ln(y) = ln(e^x +2) /x ={l’hopital says} =

[ln(e^x +2)]’ / x’ = [e^x/(e^x +2)] /1 = 1;  

♠ doing back: y = exp(z) =exp(1) = e;