What is limit x tends to 0 log(1+x)/x to the base a?

Dear Aditya,

The value tends to 1.

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. 

http://www.askiitians.com/packages/packages.aspx

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert

Just we have to expand the f(x) i.e. we can write it ltx~> 0 log a (1+x)- lt x~>o log a x

                                             So by using the expansion of log(1+x) and value of log a (0) i.e. 1 we get answer -1

Here we will use the expansion method 

Firstly lim_x-0 log_a(1+x)/x

firstly using log property we get:

lim_x-0log_a(1+x)-logx

then we change the base of log i.e lim_x-0 {log_e(1+x)/log_ea}-log_ax

then adter using expansion of log_e(1+x) we get answer " 1/log_ea

we know x→0 log(1+x)/x=1 but when the base is (e) here base is (a) so

after changing base to (e) we get the

answer   1/log a     to the base (e)

Dear Aditya,

its a very simple question.
apply the expansion of log.( log 1+x = x - x^2/2 +x^3/3 .......to infinity)
and we get ans as 1/log a to the base e.

Regards.

the limit is 1( use L''Hospital''s rule to solve)

ques :- if x tends to infinity what is d value of logx/x ?? plz solve dis step by step and also write the values if applied

Dear Student,
Please find below the solution to your problem.

apply the expansion of log.( log 1+x = x - x^2/2 +x^3/3 .......to infinity)
and we get ans as 1/log a to the base e.

Thanks and Regards