Why is the answer of  the question "lim as x→∞ of [ ln(4x+3)-ln(3x+4) ] " is ln(4/3) ??

I reached the following step: ln[ lim as x→∞ of (4x+3/3x+4) ] but can't go further on!

I really need a quick reply ... Thank You.

ln[ lim as x→∞ of (4x+3/3x+4) ]   (Note that the fraction representation is improper)

= ln[ lim as x→∞ of [ 4/3( 3x+4) -7/3]/(3x+4) ]

= ln[ lim as x→∞ of   4/3 -7/3/(3x+4) ]--------(1)

Note that the term 7/3/(3x+4) shall go to zero as x→∞ .

Hence, (1) =ln(4/3)   (shown)

Hope this helps. Peace.  


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Frederick Koh


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