# Question #c91de

Jul 14, 2016

1

#### Explanation:

assuming you mean natural log here

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

$= {\lim}_{x \to 0} \frac{x - {x}^{2} / 2 + m a t h c a l \left(O\right) \left({x}^{3}\right)}{x}$ using the Taylor expansion for the numerator about x = 0

$= {\lim}_{x \to 0} 1 + m a t h c a l \left(O\right) \left(x\right) = 1$

of you can hit it with L'Hopital as it's $\frac{0}{0}$ indeterminate so

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

you can also look at it as

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

${\lim}_{x \to 0} \ln \textcolor{red}{{\left(1 + x\right)}^{\frac{1}{x}}}$

and that bit in red is the definition of $e$ - see Bernoulli's compound interest formula, well-covered on Wiki, so you have $\ln e = 1$

you'd need a little bit more algebra to do that, ie lift the $\ln$ outside the limit, which you can do as it is continuous about the limit

Jul 14, 2016

$1$

#### Explanation:

We need to find
${\lim}_{x \to 0} \log \frac{1 + x}{x}$
Substituting $x = 0$ in the expression $\log \frac{1 + x}{x}$, we see that it is indeterminate as in the form $\frac{0}{0} .$
Therefore apply L'Hospital's Rule. Differentiate the numerator and differentiate the denominator and then take the limit.

We obtain
${\lim}_{x \to 0} \frac{\frac{1}{1 + x}}{1}$
$\implies {\lim}_{x \to 0} \frac{1}{1 + x}$
$= \frac{1}{1 + 0}$
$= 1$