lim ln[1+kx)] x-->inf.

Let us rewrite 1 as x/x

==> lim ln(x/x +kx) = lim ln[(1/x)(x)+kx

Now factor x

= lim ln x[(1/x)+k]

we know that ln xy= ln x + ln

==> lim lnx + lim ln [(1/x)+k]}

when x--> inf ==> lnx= inf

==> ln(1/x) +K = ln k

==> infinity +ln K = inf.

The limit is inf.

Also we need to know that ln(1+kx) is defined when kx> -1

To calculate the limit ln(1+kx) as x -->infinity.

Solution:

we knw that y = ln(1+kx)

y'(x) = 1/(1+kx) is > 0 for k>0 as x--> inf.

y(x) is continous and increasing unbounded, as x-->inf.

When k is zero, y = 1 fo all x or as x--> inf.

When k is negative, ln (1+kx) does not exist for negative values.

1+kx = e

We could re-write the limit above in this way:

lim ( 1/x )*ln ( 1 + kx )

Moving the ratio ( 1/x ) in front of the ln function,we could use the property of ln function:

( 1/x )*ln ( 1 + kx ) = ln ( 1+kx )^( 1/x )

we know that lim ( 1+x )^( 1/x ) , when x -> infinity is "e".

lim ln ( 1+kx )^( 1/x ) = ln lim ( 1+kx )^( 1/x )=

ln lim [( 1+kx )^( 1/kx )]^( kx*1/x ) = ln e^k = k ln e = k*1 = k