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Evaluate the limit #lim_(x rarr 0^+) (1+1/x)^x #?

Answer 1

Isaiah Allen

#lim_(xrarr0^+)(1+1/x)^x = 1#

#lim_(xrarr0^+)1/x = 0#, so

#lim_(xrarr0^+)(1+1/x) = 1#, and

#lim_(xrarr0^+)(1+1/x)^x = 1^0 = 1#

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Answer 2

Aurora Alvarado

# lim_(x rarr 0^+) (1+1/x)^x = 1 #

We seek:

# L = lim_(x rarr 0^+) (1+1/x)^x #

As the log function is monotonic we can tak logs of both sides to get:

# ln L = ln {lim_(x rarr 0^+) (1+1/x)^x} # # \ \ \ \ \ \ = lim_(x rarr 0^+) ln {(1+1/x)^x} # # \ \ \ \ \ \ = lim_(x rarr 0^+) xln (1+1/x) # # \ \ \ \ \ \ = lim_(x rarr 0^+) xln ((x+1)/x) # # \ \ \ \ \ \ = lim_(x rarr 0^+) x{ln (x+1) - lnx} # # \ \ \ \ \ \ = lim_(x rarr 0^+) {xln (x+1) - xlnx} # # \ \ \ \ \ \ = lim_(x rarr 0^+) xln (x+1) - lim_(x rarr 0^+)xlnx # # \ \ \ \ \ \ = 0 - 0 #

Thus:

# L = e^0 = 1 #

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Answer 3

Elijah Abram

The limit of (1+1/x)^x as x approaches 0 from the positive side is equal to e, where e is the mathematical constant approximately equal to 2.71828.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.