Question

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

Answer 1

`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`

Answer 2

`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`