Math Doubts

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


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


If $a$ is a constant and $x$ is a variable, then quotient of subtraction one from $a$ raised to the power of $x$ by $x$ is written as $\dfrac{a^x-1}{x}$ and the limit of this special exponential function is written mathematically as follows.

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

The limit of this special exponential function is equal to natural logarithm of $a$. It’s used as a formula in finding the limits of exponential functions in calculus.

Other forms

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

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

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


Let’s prove the limit of quotient of difference of $a^x$ and $1$ by $x$ as $x$ approaches $0$ is equal to natural logarithm of $a$.

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