Question

Nice question in calculas?

prove that

`lne=1`

Answer 1

I am nit sure it is a proof....but anyway:

Explanation:

From the definition of log:

`log_bx=a`

auch that:

`x=b^a`

that is, the integrand is equal to the base to the power of the result.

In our case we know that the Natural Logarithm, `ln`, is the one with base `e`, i.e.:

`log_e=ln`

so, using our definition of log:

`lne=1`
`log_ee=1`
so that:
`e=e^1` that is correct.

Answer 2

Show the steps below

Explanation:

show the explanation

`lne=ln[lim_(xrarroo)(1+1/x)^x]`

`=[lim_(xrarroo)ln(1+1/x)^x]`

`=[lim_(xrarroo)x*ln(1+1/x)]` product `(0*oo)`

`=[lim_(xrarroo)(ln(1+1/x)]/(1/x)]` product `(0/0)`

`=[lim_(xrarroo)(-(1/x^2)/(1+1/x)]/(-1/x^2)]=1`

Answer 3

Nearly all such proofs are cyclic.

Using Bernoulli's definition , we have:

`e = lim_(n rarr oo) (1+1/n)^n`

If we apply base `e` (Napier) logarithms to both sides:

`ln e = ln \ lim_(n rarr oo) (1+1/n)^n`

Due to the monotonicity of the logarithm function:

`ln e = lim_(n rarr oo) ln (1+1/n)^n`

Using the rules of logarithms:

`ln e = lim_(n rarr oo) n ln (1+1/n)`

We can simplify by performing a substitution:

`u=1/n iff n=1/u`, and as `n rarr oo => u rarr 0`

Thus:

`ln e = lim_(u rarr 0) 1/u ln (1+u)`

`\ \ \ \ \ = lim_(u rarr 0) (ln (1+u))/u`

This is of an indeterminate for `0//0`, so we can apply L'Hôpital's rule, to get:

`ln e = lim_(u rarr 0) (d/du ln (1+u))/(d/du u)`

`\ \ \ \ \ = lim_(u rarr 0) (1/(1+u))/(1)`

`\ \ \ \ \ = 1 \ \ \` QED