Nice question in calculas?

prove that

#lne=1#

3 Answers
Gió
Apr 28, 2018

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.

James
Apr 28, 2018

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#

Steve M
Apr 28, 2018

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

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