$x$ is a literal and the sum of the one and quotient of $4$ by $x$ is denoted as $1 + \dfrac{4}{x}$. It is raised to the power of $3x$.

${\Bigg(1+\dfrac{4}{x}\Bigg)}^{\displaystyle 3x}$

If $x$ tends to infinity, the value of the limit of the function is written as follows to express it in mathematical form.

$= \,\,\, \displaystyle \Large \lim_{x \,\to\, \infty} \normalsize {\Bigg(1+\dfrac{4}{x}\Bigg)}^{\displaystyle 3x}$

The limit of the $1$ plus $4$ by $x$ whole power of $3x$ as $x$ approaches infinity is closely matches with the following exponential limit rule.

$\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \bigg(1+\dfrac{1}{x}\bigg)^{\displaystyle x}}$ $\,=\,$ $e$

Let us try to adjust the function in the exponential notation into the above required form. In order to start the process, express the quotient of $4$ by $x$ in its reciprocal form.

$\implies$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3x}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3x}}$

The quantity in the denominator of the second term of binomial in the function should be the exponent for using this exponential limit rule. So, let’s make some adjustments at exponent position for obtaining it.

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3x \times 1}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3x \times \dfrac{4}{4}}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3 \times x \times \dfrac{4}{4}}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3 \times x \times \dfrac{4 \times 1}{4}}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3 \times x \times 4 \times \dfrac{1}{4}}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 3 \times 4 \times x \times \dfrac{1}{4}}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 12 \times \dfrac{x \times 1}{4}}}$

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle 12 \times \dfrac{x}{4}}}$

According to the power rule of exponents, the product of exponents of a quantity can be written as a power of a quantity in exponential notation.

$\,\,\,=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle \dfrac{x}{4}}\Bigg)^{\displaystyle 12}}$

The simplification process for the given function in exponential notation is completed and it is time to find its limit as the value of $x$ is closer to infinity.

$\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(\Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle \dfrac{x}{4}}\Bigg)^{\displaystyle 12}}$

Use the basic exponential limit rule to find the limit of the exponential function.

$\,\,\,=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle \dfrac{x}{4}}\Bigg)^{\displaystyle 12}}$

Suppose $y \,=\, \dfrac{x}{4}$, then convert the whole function in terms $y$.

If $x \,\to\, \infty$, then $\dfrac{x}{4} \,\to\, \dfrac{\infty}{4}$. Therefore, $\dfrac{x}{4} \,\to\, \infty$ but we have assumed that $y \,=\, \dfrac{x}{4}$. Hence, $y \,\to\, \infty$. It revealed that when the value of $x$ approaches to infinity, the value of $y$ also closer to infinity.

$\implies$ $\Bigg(\displaystyle \large \lim_{x \,\to\, \infty}{\normalsize \Bigg(1+\dfrac{1}{\Big(\dfrac{x}{4}\Big)}\Bigg)^{\displaystyle \dfrac{x}{4}}\Bigg)^{\displaystyle 12}}$ $\,=\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, \infty}{\normalsize \bigg(1+\dfrac{1}{y}\bigg)^{\displaystyle y}\Bigg)^{\displaystyle 12}}$

The expression in terms of $x$ is successfully converted into the expression in terms of $y$. It is time to evaluate the limit of the function in terms of $y$.

$\Bigg(\displaystyle \large \lim_{y \,\to\, \infty}{\normalsize \bigg(1+\dfrac{1}{y}\bigg)^{\displaystyle y}\Bigg)^{\displaystyle 12}}$

According to the limit of 1 plus 1 by x whole power x as x approaches infinity rule, the limit of $1$ plus $1$ by $y$ whole power of $y$ as $y$ closer to infinity is equal to mathematical constant $e$.

$\implies$ $\Bigg(\displaystyle \large \lim_{y \,\to\, \infty}{\normalsize \bigg(1+\dfrac{1}{y}\bigg)^{\displaystyle y}\Bigg)^{\displaystyle 12}}$ $\,=\,$ $(e)^{12}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, \infty}{\normalsize \bigg(1+\dfrac{1}{y}\bigg)^{\displaystyle y}\Bigg)^{\displaystyle 12}}$ $\,=\,$ $e^{12}$

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