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

Dear Aditya,

The value tends to 1.

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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)