Calculate lim[ln( 1 + kx)] if x -> infinite .

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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 =...

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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

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