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

In this limit problem, $x$ is a variable, which represents a real number and also an angle of the right triangle. The limit of the quotient of the exponential expression $e^x}-e^\sin{x}$ by $x-\sin{x}$ as the input $x$ approaches zero, has to evaluate in this calculus problem.

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

### Factorize the mathematical function

The given function contains exponential, trigonometric and also algebraic functions but the function is similar to the limit rule of $\dfrac{e^x}-1}{x$ as $x$ approaches $0$. So, if we eliminate $e^\sin{x}$ from the given function, then there is a possibility to convert the given function in our required form.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Big(e^x}-e^\sin{x}}\Big) \times 1}{x-\sin{x}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Big(e^x}-e^\sin{x}}\Big) \times \dfrac{e^\sin{x}}}{e^\sin{x}}}}{x-\sin{x}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{e^x}-e^\sin{x}}}{e^\sin{x}}}\Bigg) \times e^\sin{x}}}{x-\sin{x}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Bigg(\dfrac{e^x}}{e^\sin{x}}}-\dfrac{e^\sin{x}}}{e^\sin{x}}}\Bigg) \times e^\sin{x}}}{x-\sin{x}}$

Use, the quotient rule of exponents with same base rule to simplify the expression in the numerator.

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Bigg(e^x-\sin{x}}-\dfrac{\cancel{e^\sin{x}}}}{\cancel{e^\sin{x}}}}\Bigg) \times e^\sin{x}}}{x-\sin{x}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\Big(e^x-\sin{x}}-1\Big) \times e^\sin{x}}}{x-\sin{x}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\Bigg(\dfrac{e^x-\sin{x}}-1}{x-\sin{x}}\Bigg) \times \Big(e^\sin{x}}\Big)\Bigg]$

It can be further simplified by using the product rule of limits.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{e^x-\sin{x}}-1}{x-\sin{x}}\Bigg)$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Big(e^\sin{x}}\Big)$

$= \,\,\,$ $\displaystyle \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^x-\sin{x}}-1}{x-\sin{x}}\Bigg)$ $\times$ $\displaystyle \Big(\large \lim_{x \,\to\, 0}{\normalsize e^\sin{x}}\Big)$

### Evaluate the Limit of the function

Now, evaluate the limit of the $e$ raised to the power of $\sin{x}$ as $x$ approaches $0$ by the direct substitution method.

$= \,\,\,$ $\displaystyle \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^x-\sin{x}}-1}{x-\sin{x}}\Bigg)$ $\times$ $e^\sin{(0)}$

According to the sin 0 degrees, the sin of zero degrees is equal to zero.

$= \,\,\,$ $\displaystyle \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^x-\sin{x}}-1}{x-\sin{x}}\Bigg)$ $\times$ $e^{0}$

Use zero power rule of exponents to simplify the expression further.

$= \,\,\,$ $\displaystyle \Bigg(\large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^x-\sin{x}}-1}{x-\sin{x}}\Bigg)$ $\times$ $1$

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

### Evaluate the Limit of the remaining function

The given limit problem is almost simplified as the limit of $(e^x}-1)/$ as $x$ approaches $0$ formula.

Let $y = x-\sin{x}$ and convert the whole function in terms of $y$.

1. If $x \to 0$, then $\sin{x} \to \sin{(0)}$. Therefore, $\sin{x} \to 0$
2. If $x \to 0$ and $\sin{x} \to 0$, then $x-\sin{x} \to 0$. In this case, we have assumed that $y = x-\sin{x}$. Therefore $y \to 0$.

The above two steps have cleared that if $x$ approaches zero, then $y$ also tends to zero.

$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{e^x-\sin{x}}-1}{x-\sin{x}}$ $\,=\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{e^y}-1}{y}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{e^y}-1}{y}$

According to the limit of (ex-1)/x as x approaches 0 rule, the limit of $\dfrac{e^y}-1}{y$ as $y$ approaches zero is equal to one.

$= \,\,\, 1$

Therefore, it is evaluated that the limit of the ratio of $e^x}-e^\sin{x}$ to $x-\sin{x}$ as $x$ approaches zero is equal to one.

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