The limit of logarithmic function $\dfrac{\ln{(3+x)}-\ln{(3-x)}}{x}$ as $x$ approaches zero can be evaluated in two methods. One method is useful for beginners and the second method is useful for advanced calculus learners. However, the limit of ln(1+x)/x as x approaches 0 formula is used in both methods.
It is an easy method, recommendable for those who are newly learning limits.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3+x)}-\log_{e}{(3-x)}}{x}}$
The algebraic function contains a logarithmic expression in numerator and a variable in denominator. Now, let’s try to simplify this algebraic function.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(3+x)}}{x}-\dfrac{\log_{e}{(3-x)}}{x}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(3+x)}}{x}-\dfrac{\log_{e}{(3-x)}}{x}\Bigg]}$
In order to apply the limit rule of logarithmic function, the first term inside each log term should be one. It is possible by taking $3$ common from sum of terms inside each logarithmic term.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{\Bigg(3\Big(1+\dfrac{x}{3}\Big)\Bigg)}}{x}-\dfrac{\log_{e}{\Bigg(3\Big(1-\dfrac{x}{3}\Big)\Bigg)}}{x}\Bigg]}$
According to difference rule of limits, the limit of difference of two terms can be written as difference of their limits.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{\Bigg(3\Big(1+\dfrac{x}{3}\Big)\Bigg)}}{x} \Bigg]}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{\Bigg(3\Big(1-\dfrac{x}{3}\Big)\Bigg)}}{x}\Bigg]}$
Now, expand each logarithmic term by the product rule of logarithms.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(3)} + \log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x} \Bigg]}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(3)} + \log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x} \Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(3)}}{x} + \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x} \Bigg]}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(3)}}{x} + \dfrac{\log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x} \Bigg]}$
According to addition rule of logarithms, the limit of sum of terms can be written as sum of their limits.
$= \,\,\,$ $ \Bigg[ \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x} \Bigg]}$ $-$ $\Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x} \Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x}}$
$= \,\,\,$ $\require{cancel} \displaystyle \large \cancel{ \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}} }$ $-$ $\displaystyle \large \cancel{ \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3)}}{x}} }$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1-\dfrac{x}{3}\Big)}}{x}}$
Each term is almost similar to the limit rule of logarithmic function but it cannot be applied at this time. So, convert each function completely same as the limit rule of logarithmic function.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{x}}$
The denominator should be exactly same as the second term inside the logarithmic function in its numerator.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{-x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{(-x)}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{1 \times x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{1\times (-x)}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{\dfrac{3}{3} \times x}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{\dfrac{3}{3} \times (-x)}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{3 \times \dfrac{x}{3}}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{3 \times \dfrac{(-x)}{3}}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{3 \times \dfrac{x}{3}}}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{3 \times \Big(-\dfrac{x}{3}\Big)}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1}{3} \times \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{\dfrac{x}{3}}\Bigg]}$ $+$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1}{3} \times \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{\Big(-\dfrac{x}{3}\Big)}\Bigg]}$
According to constant multiple rule of limits, the limit of product of a constant and a function can be written as product of constant and limit of the function.
$= \,\,\,$ $\dfrac{1}{3} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{x}{3}\Big)}}{\dfrac{x}{3}}}$ $+$ $\dfrac{1}{3} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\Big(-\dfrac{x}{3}\Big)\Big)}}{\Big(-\dfrac{x}{3}\Big)}}$
Take $y = \dfrac{x}{3}$. If $x \to 0$, then $\dfrac{x}{3} \to \dfrac{0}{3}$. So, $\dfrac{x}{3} \to 0$ but $\dfrac{x}{3}$ is $y$. Therefore, $y \to 0$. It proved that if $x$ approaches zero, then $y$ also approaches zero.
Similarly, take $z = -\dfrac{x}{3}$. If $x \to 0$, then $\dfrac{x}{3} \to \dfrac{0}{3}$. So, $\dfrac{x}{3} \to 0$ and $-\dfrac{x}{3} \to 0$ but $-\dfrac{x}{3}$ is $z$. Therefore, $z \to 0$. It also proved that if $x$ approaches zero, then $z$ also approaches zero.
$= \,\,\,$ $\dfrac{1}{3} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+y)}}{y}}$ $+$ $\dfrac{1}{3} \times \displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+z)}}{z}}$
$= \,\,\,$ $\dfrac{1}{3} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+y)}}{y}}$ $+$ $\dfrac{1}{3} \times \displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+z)}}{z}}$
Find the limit of each logarithmic function as per limit rule of logarithmic function.
$= \,\,\,$ $\dfrac{1}{3} \times 1$ $+$ $\dfrac{1}{3} \times 1$
$= \,\,\,$ $\dfrac{1}{3}+\dfrac{1}{3}$
$= \,\,\,$ $\dfrac{2}{3}$
It is an advanced method, which can be understood by those who already have good knowledge on limits.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(3+x)}-\log_{e}{(3-x)}}{x}}$
In numerator, two logarithmic terms are in subtraction form and they can be combined as a logarithmic term by using quotient rule of logarithms.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{3+x}{3-x}\Big)}}{x}}$
Try to convert the logarithmic term in the numerator same as the limit rule of logarithmic function. In order to convert it, add and subtract $x$ in the numerator of the logarithmic function.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{3-x+x+x}{3-x}\Big)}}{x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{3-x+2x}{3-x}\Big)}}{x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{3-x}{3-x}+\dfrac{2x}{3-x}\Big)}}{x}}$
$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(\dfrac{\cancel{3-x}}{\cancel{3-x}}+\dfrac{2x}{3-x}\Big)}}{x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{2x}{3-x}\Big)}}{x}}$
Now, the function is almost similar to the limit rule of logarithmic function.
Take $q = \dfrac{2x}{3-x}$ and express the variable $x$ in terms of $q$.
$\implies q(3-x) = 2x$
$\implies q \times 3 -q \times x = 2x$
$\implies 3q -qx = 2x$
$\implies 3q = qx+2x$
$\implies 3q = (q+2)x$
$\implies \dfrac{3q}{q+2} = x$
$\,\,\, \therefore \,\,\,\,\,\,$ $x = \dfrac{3q}{q+2}$
Therefore, convert the entire function in terms of $q$ from $x$ by replacing $\dfrac{2x}{3-x}$ by $q$ and $x$ by $\dfrac{3q}{q+2}$.
According to $q = \dfrac{2x}{3-x}$, then $q = \dfrac{2}{\dfrac{3-x}{x}}$. Now, $q = \dfrac{2}{\dfrac{3}{x}-\dfrac{x}{x}}$. Therefore $q = \dfrac{2}{\dfrac{3}{x}-1}$ and it is used to find the input value for the limiting operation.
If $x \to 0$, then $\dfrac{3}{x} \to \dfrac{3}{0}$. So, $\dfrac{3}{x} \to \infty$. Now, $\dfrac{3}{x}-1 \to \infty -1$. Therefore, $\dfrac{3}{x}-1 \to \infty$. Similarly, $\dfrac{2}{\dfrac{3}{x}-1} \to \dfrac{1}{\infty}$. Therefore, $\dfrac{2}{\dfrac{3}{x}-1} \to 0$ but $\dfrac{2}{\dfrac{3}{x}-1}$ is $q$. So, $q \to 0$
It is proved that If $x \to 0$, then $q \to 0$. It means if $x$ approaches zero then $q$ also approaches zero.
Now, transform the entire function in terms of $q$ from $x$ as per this transforming equivalents.
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Big(1+\dfrac{2x}{3-x}\Big)}}{x}}$ $\,=\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+q)}}{\dfrac{3q}{q+2}}}$
$\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+q)}}{\dfrac{3q}{q+2}}}$
The transformed function can be further simplified to evaluate the limit of this algebraic function.
$= \,\,\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+q)}}{\dfrac{3 q}{q+2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+q)}}{\dfrac{3 \times q}{q+2}}}$
$= \,\,\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \Bigg[\dfrac{\log_{e}{(1+q)}}{q} \times \dfrac{1}{\dfrac{3}{q+2}}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \Bigg[ \dfrac{\log_{e}{(1+q)}}{q} \times \dfrac{q+2}{3} \Bigg]}$
Now, use product rule of limits to express limit of product of functions as product of their limits.
$= \,\,\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+q)}}{q}}$ $\times$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{q+2}{3}}$
$\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+q)}}{q}}$ $\times$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{q+2}{3}}$
Now, evaluate the limit of the function by the limit of logarithmic function rule.
$= \,\,\,$ $1 \times \displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{q+2}{3}}$
$= \,\,\,$ $\displaystyle \large \lim_{q \,\to\, 0}{\normalsize \dfrac{q+2}{3}}$
Now, evaluate the limit of algebraic function by the direct substitution method.
$= \,\,\,$ $\dfrac{(0)+2}{3}$
$= \,\,\,$ $\dfrac{0+2}{3}$
$= \,\,\,$ $\dfrac{2}{3}$
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