Math Doubts

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


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

$a$ and $x$ are two literal numbers but $a$ represents a constant and $x$ represents a variable. The ratio of subtraction of one from the $a$ raised to the power of $x$ to literal number $x$ is written as follows.

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

The value of this function is requisite when the value of $x$ tends to zero. It is expressed in limit by the calculus in mathematical form in the following way.

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


Learn how to derive limit of $\dfrac{a^{\displaystyle \normalsize x}-1}{x}$ as $x$ approaches $0$ in calculus.

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