Math Doubts

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 $y$.

  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.

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

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

According to the limit of (ex-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.

Math Doubts
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more