The limit of $e$ raised to the power $3$ plus $x$ minus $\sin{x}$ minus $e$ raised to the power $3$ by $x$ as $x$ approaches to $0$ is solved in fundamental limit method and it will be solved here by L’Hospital’s Rule (or L’Hopital’s rule)

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^{3\displaystyle +x}-\sin{x}-e^3}{x}}$

Let’s get started to learn how to find the limit of the given rational function by L’Hopital’s Rule (or L’Hospital’s Rule).

Let us try to find the limit of the given rational function by using direct substitution method as the value of $x$ approaches to $0$.

$=\,\,\,$ $\dfrac{e^{3 \displaystyle + \small 0}-\sin{(0)}-e^3}{0}$

$=\,\,\,$ $\dfrac{e^3-0-e^3}{0}$

The sine of angle zero radian is equal to zero as per the trigonometric mathematics.

$=\,\,\,$ $\dfrac{e^3-e^3}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

It is evaluated that the limit of the function is indeterminate. Hence, it is better to think about an alternative mathematical approach to find its limit.

The given function is eligible for evaluating its limit by the L’Hopital’s Rule (or L’Hospital’s Rule). Now, differentiate the expressions in both numerator and denominator with respect to $x$.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\Big(e^{3\large +\Large x}-\sin{x}-e^3\Big)}{\dfrac{d}{dx}{(x)}}}$

The derivative of the difference of the terms can be evaluated as per the difference rule of the differentiation.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\big(e^{3\large +\Large x}\big)-\dfrac{d}{dx}{\sin{x}}-\dfrac{d}{dx}{(e^3)}}{\dfrac{d}{dx}{(x)}}}$

The $e$ raised to the power $3$ plus $x$ can be split as a product of two quantities in exponential notation as per the product rule of the exponents.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}\big(e^3 \times e^{\Large x}\big)-\dfrac{d}{dx}{\sin{x}}-\dfrac{d}{dx}{(e^3)}}{\dfrac{d}{dx}{(x)}}}$

The factor $e$ raised to the power $3$vcolkfaZZ is a constant. Hence, it can be separated from the differentiation by using the constant multiple rule of the limits.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^3 \times \dfrac{d}{dx}\big(e^{\Large x}\big)-\dfrac{d}{dx}{\sin{x}}-\dfrac{d}{dx}{(e^3)}}{\dfrac{d}{dx}{(x)}}}$

Use derivative rule of natural exponential function, derivative rule of sine function and derivative rule of a constant to find the derivatives in the numerator. Similarly, find the derivative of the variable by using the derivative rule of a variable.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^3 \times e^{\Large x}-\cos{x}-0}{1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \big(e^3 \times e^{\Large x}-\cos{x}\big)}$

Now, use the direct substitution method one more time to find the limit of the function as the value of $x$ approaches to zero.

$=\,\,\,$ $e^3 \times e^0-\cos{(0)}$

According to the zero power rule, the mathematical constant $e$ raised to the power zero is equal to one. The cosine of angle zero radian is equal to one as per the trigonometry.

$=\,\,\,$ $e^3 \times 1-1$

$=\,\,\,$ $e^3-1$

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