Lim x->0 [log(1+ax)+log(1+bx)]/x =?
#Lim_(x->0)[log_e(1+ax)+log_e(1+bx)]/x =?#
3 Answers
Explanation:
Starting from the fundamental limit:
we have:
and then:
Explanation:
We will use the limit
I didn't use
to emphasize even more that fact. Anyway, on with the computation: let's break the fraction in the sum of two fractions:
We now multiply each fraction by
the expression becomes
As you can see, we have the same quantities summed inside the logarithm, and at the denominator, and they are tending to zero, since
Explanation:
We use the Standard Limit :
is continuous,