Limits Of Basic Logarithmic Functions


I.  $\lim\limits_{x \rightarrow 0} \dfrac{\log_a(1 + x)}{x} = \log_a e$:

$\lim\limits_{x \rightarrow 0} \dfrac{\log_a(1 + x)}{x}$
$= \lim\limits_{x \rightarrow 0} \dfrac{1}{x} \times \log_a(1 + x)$
$= \lim\limits_{x \rightarrow 0}  \log_a(1 + x)^{\frac{1}{x}}$

Let $x = \dfrac{1}{y}$
$\therefore$ As $x \rightarrow 0$, $y \rightarrow \infty$
Therefore, the above limit becomes:
$\lim\limits_{y \rightarrow \infty}  \log_a\left(1 + \dfrac{1}{y}\right)^{y} = \log_a\left[\lim\limits_{y \rightarrow \infty}\left(1 + \dfrac{1}{y}\right)^{y} \right]$     (We will see the validity of this step in a while)

We know from , that $\lim\limits_{y \rightarrow \infty}\left(1 + \dfrac{1}{y}\right)^{y} = e$, therefore, the above limit becomes:

$= \log_a e$

-----------book page break-----------
If $\lim\limits_{x \rightarrow a} g(x) = L$ exists and $f(x)$ is continuous at $x = L$, then,

$\lim\limits_{x \rightarrow a} f(g(x)) = f(\lim\limits_{x \rightarrow a} g(x))$, 

Since $log_a x$ is continuous at $x = e$ our substitution in the previous step is valid.

II.  $\lim\limits_{x \rightarrow 0} \dfrac{\ln (1 + x)}{x} = 1$:
This one is a direct consequence of the proof in the previous section. 

$\lim\limits_{x \rightarrow 0} \dfrac{\ln (1 + x)}{x}$

$= \lim\limits_{x \rightarrow 0} \dfrac{\log_e (1 + x)}{x}$

$= \log_e e = 1$ 

III.  $\lim\limits_{x \rightarrow 0} \dfrac{(a^x - 1)}{x} = \ln a$:
Let's assume $\lim\limits_{x \rightarrow 0} \dfrac{(a^x - 1)}{x} = L$ where $L$ is a finite number.
Let $x = \dfrac{1}{y} \therefore$ as $x \rightarrow 0$ $y \rightarrow \infty$
Making the above substitution we get:
$\lim\limits_{y \rightarrow \infty} \dfrac{(a^{\frac{1}{y}} - 1)}{\dfrac{1}{y}} = L$

-----------book page break-----------
$\Rightarrow \lim\limits_{y \rightarrow \infty} (a^{\frac{1}{y}} - 1) = \lim\limits_{y \rightarrow \infty} L \times \dfrac{1}{y}$

$\Rightarrow \lim\limits_{y \rightarrow \infty} \left(a^{\frac{1}{y}}\right) = \lim\limits_{y \rightarrow \infty}\left( \dfrac{L}{y} + 1\right)$

$\Rightarrow \lim\limits_{y \rightarrow \infty} \left(a^{\frac{1}{y}}\right)^y = \lim\limits_{y \rightarrow \infty}\left( \dfrac{L}{y} + 1\right)^y$

$\Rightarrow \lim\limits_{y \rightarrow \infty} a = \lim\limits_{y \rightarrow \infty}\left( \dfrac{L}{y} + 1\right)^{L \times \frac{y}{L}}$

Substituting $\dfrac{1}{z} = \dfrac{L}{y}$ in the above equation, we get:
 
$a = \lim\limits_{z \rightarrow \infty}\left( \dfrac{1}{z} + 1\right)^{L \times z}$

$\Rightarrow a = \left[ \lim\limits_{z \rightarrow \infty}\left( \dfrac{1}{z} + 1\right)^{z} \right]^L$

$\Rightarrow a = e^L$

$\therefore L = \log_e a = \ln a$

Therefore, $\lim\limits_{x \rightarrow 0} \dfrac{(a^x - 1)}{x} = \ln a$