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$\displaystyle \lim_{x \,\to\, 0}{\dfrac{a^x-1}{x}}$ formula


$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle \normalsize x}-1}{x}}$ $\,=\,$ $\log_{e}{a}$


Let $a$ be a constant and $x$ be a variable. $a$ raised to the power of $x$ forms an exponential function $a^{\displaystyle \normalsize x}$. The subtraction of one from the exponential function forms a mathematical expression $a^{\displaystyle \normalsize x}-1$. The quotient of the function in difference form by the variable $x$ is written mathematically as follows.

$\dfrac{a^{\displaystyle \normalsize x}-1}{x}$

The limit of this type of special mathematical function has to evaluate In calculus as the value of variable is closer to zero. In this case, the literal $x$ represents a variable. Therefore, the limit of this function as the value of $x$ approaches $0$ is written as follows.

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle \normalsize x}-1}{x}}$

The limit of the quotient of $x^{th}$ power of $a$ minus $1$ by $x$ as the value of $x$ tends to $0$ is equal to natural logarithm of constant $a$.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle \normalsize x}-1}{x}}$ $\,=\,$ $\log_{e}{a}$

According to the natural logarithms, the logarithm of a to base neper constant is also simply written as natural logarithm of a.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle \normalsize x}-1}{x}}$ $\,=\,$ $\ln{a}$

This mathematical equation is used as a formula in calculus to find the limit of functions in which the exponential functions are involved.

Other forms

This standard result in limits can be written in several ways in calculus.

$(1). \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{c^{\displaystyle \normalsize y}-1}{y}}$ $\,=\,$ $\ln{(c)}$

$(2). \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{m^{\displaystyle \normalsize u}-1}{u}}$ $\,=\,$ $\log_{e}{(m)}$

$(3). \,\,\,$ $\displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{l^{\displaystyle \normalsize z}-1}{z}}$ $\,=\,$ $\ln{(l)}$


Learn how to derive the limit rule for the quotient of $a$ to the $x$th power minus $1$ by $x$ as $x$ approaches $0$.

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