# Lim x->0 [log(1+ax)+log(1+bx)]/x =?

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#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,**