Question

How do you find the limit of $\left( {\dfrac{{x + \sin x}}{x}} \right)$ as $x$ approaches $0$?

Answer 1

In this question we need to find the limit $\left( {\dfrac{{x + \sin x}}{x}} \right)$ as $x$ approaches $0$. To find the limit we will use the identity $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$. The limit of $\left( {\dfrac{{x + \sin x}}{x}} \right)$ as $x$ approaches $0$ is the limit of the form $\dfrac{0}{0}$ when we put $0$ in the given limit in the place of$x$. So we will first convert this limit so that it is not of form$\dfrac{0}{0}$.

Complete step by step solution:
Let us try to solve this problem in this question. We need to find the limit of $\left( {\dfrac{{x + \sin x}}{x}} \right)$ as $x$approaches $0$. As we put the value of $x = 0$since$x \to 0$.
We get the value of the limit equals to $\infty$.
We will first simplify this limit, so that we will get a finite limit. We will use the algebraic properties of limit. For finding this limit we will use addition property of limits which says that Limit of sum of two functions is sum of the limits of the functions, i.e.,
$\mathop {\lim }\limits_{x \to a} [f(x) + g(x)] = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)$
By using the above property of limits we can write our given limit as,
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{x + \sin x}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{x}{x} + \dfrac{{\sin x}}{x}} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( {1 + \dfrac{{\sin x}}{x}} \right)$
Now, we will apply the limit of the sum of two functions property, since we that 1 is a constant function the limit of a constant function always exists. So our limit will become by using limit of two functions
$\mathop {\lim }\limits_{x \to 0} \left( {1 + \dfrac{{\sin x}}{x}} \right) = \mathop {\lim }\limits_{x \to 0} (1) + \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right)$
As we know $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{\sin x}}{x}} \right) = 1$ we got, and limit of constant function is constant the limit of
$\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{x + \sin x}}{x}} \right) = 1 + 1 \\\ \mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{x + \sin x}}{x}} \right) = 2 \\\$
Hence we get the limit of $\left( {\dfrac{{x + \sin x}}{x}} \right)$ as $x$ approaches$0$ is 2.

Note: We can also find the limit of this function by also using L’Hospital rule because the given limit is of the form$\dfrac{0}{0}$. Also we can find the limit of this function by the definition of limit, but that could be lengthy and difficult to grasp.