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 $e^{\displaystyle x}-e^{\displaystyle \sin{x}}$ by $x-\sin{x}$ as the input $x$ approaches zero, has to evaluate in this calculus problem.

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

Factorize the mathematical function

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

Evaluate the Limit of the remaining function

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

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

If $x \to 0$, then $\sin{x} \to \sin{(0)}$. Therefore, $\sin{x} \to 0$
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.

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

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

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

$= \,\,\, 1$

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