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

2 Answers
Jul 23, 2017

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

Explanation:

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

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

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

Jul 23, 2017

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

Explanation:

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