Question

Show that lim h->0 (1+h)^(1/h)=e

Hint: Take ln of the quantity in the limit and note that, by the definition of the derivative, (lnx) = lim h->0 ((ln(1+h)-ln1)/(h))

Answer 1

this is pretty much one definition of e

Answer 2

but if you want some work, first take the log get
\[\frac{1}{h}\ln(1+h)\] now take limit as h goes to zero. this is the definition of the derivative of the log of 1
the derivative of
\[ln(x)\]
is
\[\frac{1}{x}\] and the derivative at 1 is obviously 1,
so this limit is 1.
since the limit of the logarithm is 1, then the original limit must be
\[e^1=e\]

Answer 3

You make it look so simple. Thank you for sharing you intelligence!

Answer 4

practice not intelligence, but you are welcome