The algebraic, natural exponential and trigonometric functions are involved to form a special rational function, and we have to evaluate this function as $x$ approaches $0$ in the given limit problem.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(e^{-3x+2}-e^2)\sin{\pi x}}{4x^2}}$
In numerator, there are two exponential terms in the first factor. The exponent of the first exponential term is an algebraic expression, which can be split as product of two exponents with same base by the product rule of exponents.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(e^{-3x} \times e^2-e^2)\sin{\pi x}}{4x^2}}$
There is a common factor $e^2$ in the numerator and it can be separated by taking out the common factor from them.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(e^{-3x} \times e^2-1 \times e^2)\sin{\pi x}}{4x^2}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^2 \times (e^{-3x}-1)\sin{\pi x}}{4x^2}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^2 \times (e^{-3x}-1)\sin{\pi x}}{4x^2}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^2 \times (e^{-3x}-1)\sin{\pi x}}{4 \times x^2}}$
The factor $e^2$ is a constant in the expression of the numerator and $4$ is also a constant in the denominator. So, we can separate them from the expression mathematically.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{e^2}{4} \times \dfrac{ (e^{-3x}-1)\sin{\pi x}}{x^2}\Bigg)}$
The constant factor can be taken out from the expression by using the constant multiple rule of limits.
$=\,\,\,$ $\dfrac{e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{ (e^{-3x}-1)\sin{\pi x}}{x^2}}$
The rational expression expresses the combination of the limit rule of exponential function and a limit rule of trigonometric function. Hence, we should split the function as two factors and then we can evaluate it easily by the corresponding formulas.
$=\,\,\,$ $\dfrac{e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{ (e^{-3x}-1)\sin{\pi x}}{x \times x}}$
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{(e^{-3x}-1)\sin{\pi x}}{x \times x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{(e^{-3x}-1)}{x} \times \dfrac{\sin{\pi x}}{x}\Bigg)}$
Now, use the product rule of limits for evaluating the limit of the product of functions by the product of their limits.
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{(e^{-3x}-1)}{x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
We can now evaluate the limit of each function. So, let us find the limit of exponential function first and then we will evaluate the limit of the trigonometric function.
The first limit function is almost similar to the limit rule of an exponential function. The exponent of the natural exponential function is $-3x$. So, the expression in the denominator and input of the limiting operation should also be $-3x$. Now, let us make some adjustments in the exponential function for transforming it into required form.
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \Bigg[1 \times \dfrac{(e^{-3x}-1)}{x}\Bigg]}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{-3}{-3} \times \dfrac{(e^{-3x}-1)}{x}\Bigg]}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \Bigg[(-3) \times \dfrac{(e^{-3x}-1)}{-3 \times x}\Bigg]}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times \Bigg((-3) \times \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{-3x}-1}{-3x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{e^2}{4} \times (-3) \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{-3x}-1}{-3x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{(-3) \times e^2}{4} \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{-3x}-1}{-3x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\displaystyle \dfrac{-3e^2}{4} \times \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{-3x}-1}{-3x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
When $x \,\to\, 0$, then $-3 \times x \,\to\, -3 \times 0$. Therefore $-3x \,\to\, 0$.
$=\,\,\,$ $\displaystyle \dfrac{-3e^2}{4} \times \Bigg(\large \lim_{-3x \,\to\, 0}{\normalsize \dfrac{e^{-3x}-1}{-3x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
Assume, $y = -3x$ and express the first limit function in terms of $y$ but keep the second function as it is.
$=\,\,\,$ $\displaystyle \dfrac{-3e^2}{4} \times \Bigg(\large \lim_{y \,\to\, 0}{\normalsize \dfrac{e^y-1}{y}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
According to the limit rule of e^x-1 by x as x approaches 0, the limit of $\dfrac{e^y-1}{y}$ as $y$ approaches $0$ is equal to one.
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \Bigg(1 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{x}}$
Now, the limit of rational expression can be evaluated by the limit rule of trigonometric function. Actually, the current rational function is almost similar to the limit rule of trigonometric function and we have to take some steps to get it in required form.
In this case, the denominator should be $\pi x$ and the input of the limiting operation should also be same for applying the limit rule of trigonometric function in sine function.
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(1 \times \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\pi}{\pi} \times \dfrac{\sin{\pi x}}{x}\Bigg)}$
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\pi \times \dfrac{\sin{\pi x}}{\pi \times x}\Bigg)}$
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\pi \times \dfrac{\sin{\pi x}}{\pi x}\Bigg)}$
$=\,\,\,$ $\dfrac{-3e^2}{4} \times \pi \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{\pi x}}$
$=\,\,\,$ $\dfrac{-3e^2 \times \pi}{4} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{\pi x}}$
Let $x \,\to\, 0$, then $\pi \times x \,\to\, \pi \times 0$. Therefore, $\pi x \,\to\, \0$. It is proved that $\pi x$ also approaches zero when $x$ approaches zero.
$=\,\,\,$ $\dfrac{-3e^2 \times \pi}{4} \times \displaystyle \large \lim_{\pi x \,\to\, 0}{\normalsize \dfrac{\sin{\pi x}}{\pi x}}$
Assume $z = \pi x$. Now, express the function in terms of $z$ for evaluating the limit of the function.
$=\,\,\,$ $\dfrac{-3e^2 \times \pi}{4} \times \displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{\sin{z}}{z}}$
As per the limit of sinx/x as x approaches 0 formula, the limit of the $\dfrac{\sin{z}}{z}$ as $z$ tends to zero is equal to one.
$=\,\,\,$ $\dfrac{-3\pi e^2}{4} \times 1$
$=\,\,\,$ $\dfrac{-3\pi e^2}{4}$
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