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Find the limit as x approaches infinity of #y=ln( 2x )-ln(1+x)#?

Answer 1

#ln2#

We have #lim_(xrarroo)(ln(2x)-ln(1+x))#

Since #lna-lnb=ln(a/b)#, we write:

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

According to the limit chain rule, let's say we have a function #(f@g)#, or #f(g(x))#. The limit of this function, when #x# tends to #a#, is given by #L#. To find #L#, we first find #lim_(xrarra)g(x)#. The answer to this is given as #b#. Then, we find #lim_(urarrb)f(u)#, where #u=g(x)#. The answer for this final limit is #L#.

Here we take #f(x)=lnx# and #g(x)=(2x)/(1+x)#. Also, here, #a=oo#. First we take:

#lim_(xrarroo)(2x)/(1+x)#

#2lim_(xrarroo)x/(1+x)#

We input #oo#:

#2(oo/oo)#

We see that we get an indeterminate form. According to L'Hopital's Rule, when #lim_(xrarrc)f(x)/g(x)# yields an indeterminate form, such as #oo/oo#, the limit is given by #lim_(xrarrc)(f'(x))/(g'(x))#.

We go back to #lim_(xrarroo)x/(1+x)#. Since #d/dxx=1# and #d/dx(1+x)=1#, we write it as:

#2*(1/1)#

#2(1)#

#=2#

So we say that #b=2#. We compute:

#lim_(urarr2)ln(u)#

Simply input and we get:

#ln(2)#

Our answer.

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Answer 2

William Abdalla

The limit as x approaches infinity of y=ln(2x)-ln(1+x) is equal to ln(2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.