How do you find the limit of #ln(2x)-ln(1+x)# as x approaches infinity?

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Answer 1 · 2017-01-26T23:30:09

Use the property that the difference to two logarithms is division within argument.
Use L'Hôpital's rule on argument.

Given: #lim_(xtooo)(ln(2x) - ln(x + 1))#

Use the property that the difference to two logarithms is division within argument.

#lim_(xtooo)ln((2x)/(x + 1))#

#ln(lim_(xtooo)(2x)/(x + 1))#

Compute the derivative of the numerator:

#(d(2x))/dx = 2#

Compute the derivative of the denominator:

#(d(x+ 1))/dx = 1#

The new expression is:

#ln(lim_(xtooo)(2/1)) = ln(2)#

Therefore, the original limit is the same:

#lim_(xtooo)(ln(2x) - ln(x + 1)) = ln(2)#