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

2 Answers
Jim H
May 2, 2016

#lim_(xrarroo)(1+2/x)^x = e^2#

Explanation:

Use #lim_(urarroo)(1+1/u)^u = e#

#lim_(xrarroo)(1+2/x)^x = lim_(xrarroo)(1+1/(x/2))^x#

# = lim_(xrarroo)((1+1/(x/2))^(x/2))^2#

# = (lim_(xrarroo)(1+1/(x/2))^(x/2))^2#

Now we have the form above with #u = x/2#, so we evaluate the limit.

# = e^2#

Ken Rubenstein
May 4, 2016

Jim, awesome ..

What this limit really represents is essentially the horizontal asymptote y = #e^2#, reflecting the function's long term graphical behavior.

Explanation:

Here are a couple of TI screenshots showing the graph and the decimal expansion for #e^2#.

enter image source here

enter image source here
If we went even further out to the right and then asked some random guy on the street if they are looking at a straight line, they would say "yes!".
enter image source here

But in fact you are looking at the curve endowed with concavity, not a straight line. The curve is asymptotically approaching the value of #y=e^2#

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