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

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

Jun 18, 2018

${\lim}_{x \to 0} \frac{\ln \left(1 + a x\right) + \ln \left(1 + b x\right)}{x} = a + b$

#### Explanation:

Starting from the fundamental limit:

${\lim}_{x \to 0} \ln \frac{1 + x}{x} = 1$

we have:

${\lim}_{x \to 0} \ln \frac{1 + a x}{x} = a {\lim}_{x \to 0} \ln \frac{1 + a x}{a x} = a$

and then:

${\lim}_{x \to 0} \frac{\ln \left(1 + a x\right) + \ln \left(1 + b x\right)}{x} = a + b$

Jun 18, 2018

$a + b$

#### Explanation:

We will use the limit

${\lim}_{k \setminus \to 0} \setminus \frac{\setminus \ln \left(1 + k\right)}{k} = 1$

I didn't use $x$ on purpose, since I want to emphasize the idea behind this exercise: is is important that you have the same quantity inside the logarithm (after the "one plus") and at the denominator, no matter what that expression is, as long as it tends to zero. Some may even prefer writing

${\lim}_{f \left(x\right) \setminus \to 0} \setminus \frac{\setminus \ln \left(1 + f \left(x\right)\right)}{f \left(x\right)} = 1$

to emphasize even more that fact. Anyway, on with the computation: let's break the fraction in the sum of two fractions:

$\setminus \frac{\ln \left(1 + a x\right) + \log \left(1 + b x\right)}{x} = \setminus \frac{\ln \left(1 + a x\right)}{x} + \setminus \frac{\log \left(1 + b x\right)}{x}$

We now multiply each fraction by $1$, using the obvious equalities

$1 = \frac{a}{a} = \frac{b}{b}$

the expression becomes

$a \setminus \frac{\ln \left(1 + \textcolor{red}{a x}\right)}{\textcolor{red}{a x}} + b \setminus \frac{\log \left(1 + \textcolor{red}{b x}\right)}{\textcolor{red}{b x}}$

As you can see, we have the same quantities summed inside the logarithm, and at the denominator, and they are tending to zero, since $x \setminus \to 0$. This means that both ratios tend to $1$, and thus the whole expression tends to

$a \cdot 1 + b \cdot 1 = a + b$

Jun 18, 2018

$a + b$.

#### Explanation:

We use the Standard Limit : ${\lim}_{t \to 0} {\left(1 + t\right)}^{\frac{1}{t}} = e$.

$\text{The Limit} = {\lim}_{x \to 0} \frac{{\log}_{e} \left(1 + a x\right) + {\log}_{e} \left(1 + b x\right)}{x}$,

$= \lim \left[\frac{1}{x} {\log}_{e} \left(1 + a x\right) + \frac{1}{x} {\log}_{e} \left(1 + b x\right)\right]$,

$= \lim \left[{\log}_{e} {\left(1 + a x\right)}^{\frac{1}{x}} + {\log}_{e} {\left(1 + b x\right)}^{\frac{1}{x}}\right]$,

$= \lim \left[{\log}_{e} {\left\{{\left(1 + a x\right)}^{\frac{1}{a x}}\right\}}^{a} + {\log}_{e} {\left\{{\left(1 + b x\right)}^{\frac{1}{b x}}\right\}}^{b}\right]$,

$= \lim \left[a {\log}_{e} \left\{{\left(1 + a x\right)}^{\frac{1}{a x}}\right\} + b {\log}_{e} \left\{{\left(1 + b x\right)}^{\frac{1}{b x}}\right\}\right]$,

$= a \cdot {\lim}_{x \to 0} {\log}_{e} \left\{{\left(1 + a x\right)}^{\frac{1}{a x}}\right\} + b \cdot {\lim}_{x \to 0} {\log}_{e} \left\{{\left(1 + b x\right)}^{\frac{1}{b x}}\right\}$,

$\text{Now, "lim_(x to 0){(1+ax)^(1/(ax))}=e," and, as "log_e" function}$

is continuous,

$\therefore {\lim}_{x \to 0} {\log}_{e} \left\{{\left(1 + a x\right)}^{\frac{1}{a x}}\right\}$

$= {\log}_{e} \left[{\lim}_{x \to 0} \left\{{\left(1 + a x\right)}^{\frac{1}{a x}}\right\}\right]$,

$= {\log}_{e} e$

$= 1$.

$\text{Thus, the Limit} = a \cdot 1 + b \cdot 1 = a + b$.