The sine of angle $x$ is subtracted from $e$ raised to the power $3$ plus $x$, and $e$ raised to the power $3$ is subtracted from the difference of the terms. The limit of the difference by the algebraic function $x$ should be evaluated as the value of $x$ approaches to zero in this limit problem.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{3\displaystyle+x}-\sin{x}-e^3}{x}}$
In this limit problem, both exponential form functions, trigonometric function and an algebraic function are involved in forming the rational function. Let us learn how to find the limit of the given rational expression as the value of $x$ tends to zero.
There are two exponential form functions and a trigonometric function sine in the numerator and an algebraic function in the denominator. In limits, there is no formula that contains both exponential and trigonometric functions. Hence, it is better to separate them for moving ahead in finding the limit of the given function.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{3\displaystyle+x}-e^3-\sin{x}}{x}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{e^{3\displaystyle+x}-e^3}{x}-\dfrac{\sin{x}}{x}\Bigg)}$
The limit of difference of any two functions can be calculated by the difference of their limits as per the difference rule of the limits.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{3\displaystyle+x}-e^3}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$
The second term represents the trigonometric limit rule in sine function exactly. So, let’s concentrate on the first term in the expression. In this term, there is a constant in exponential notation in both terms of the numerator. So, split the first term by using product rule of exponents.
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^3 \times e^{\displaystyle x}-e^3}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$
$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^3 \times e^{\displaystyle x}-e^3 \times 1}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$
Now, take out the common factor from both terms in the numerator of the first term.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^3 \times (e^{\displaystyle x}-1)}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \bigg(e^3 \times \dfrac{e^{\displaystyle x}-1}{x}\bigg)}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$
The constant factor can be separated from the limiting operation by the constant multiple rule of the limits.
$= \,\,\,$ $e^3 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^{\displaystyle x}-1}{x}}$ $-$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}$
The second factor in the first term represents the natural exponential limit rule. Therefore, the limit of the $e$ raised to the power $x$ minus one by $x$ as $x$ approaches to $0$ is equal to one. According to the trigonometric limit rule in sine function, the limit of the sin of angle $x$ by $x$ as $x$ is closer to $0$ is equal to one.
$= \,\,\,$ $e^3 \times 1-1$
$= \,\,\,$ $e^3-1$
$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{e^{3\displaystyle +x}-\sin{x}-e^3}{x}}$
Learn how to calculate the limit of the $e$ raised to the power $3$ plus $x$ minus sine of angle $x$ minus $e$ cubed by $x$ as $x$ approaches zero by using L’Hospital’s rule.
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