
How do you find the limit of \[\dfrac{{\sin x}}{x}\] as x approaches infinity?
Answer
556.5k+ views
Hint: Here in this question we need to find the limit for the function. Let we define the given function as \[f(x)\] and we apply the limit to it. The function will get closer and closer to some number. When the x approaches to infinity we need to find the limit of a function.
Complete step-by-step answer:
The idea of a limit is a basis of all calculus. The limit of a function is defined as let \[f(x)\]be a function defined on an interval that contains\[x = a\]. Then we say that\[\mathop {\lim }\limits_{x \to a} f(x) = L\], if for every \[\varepsilon > 0\]there is some number \[\delta > 0\]such that \[\left| {f(x) - L} \right| < \varepsilon \]whenever\[0 < \left| {x - a} \right| < \delta \].
Here we have to find the value of the limit when x is infinity. The function is trigonometry. Let we define the given function as\[f(x) = \dfrac{{\sin x}}{x}\]. Now we are going to apply the limit to the function \[f(x)\]so we have
\[\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin x}}{x}\]
The function \[f(x)\] is a trigonometric function. As x approaches \[\infty \] we have to substitute the x as\[\infty \].
When we consider the x value we will consider it as infinity. So by applying limit we have
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{{\sin (\infty )}}{\infty }\]
The sine function of infinity is a finite value and the any number divided by infinity is zero so we have
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } f(x) = 0\]
Hence the when the x approaches to zero for the function \[\dfrac{{\sin x}}{x}\]
Therefore we have found the limit for the function \[f(x) = \dfrac{{\sin x}}{x}\]
Hence \[\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin x}}{x} = 0\]
So, the correct answer is “0”.
Note: The function is of trigonometry and it is in form of fraction. The sine function of infinity is some finite value. When the x approaches to infinity, it implies that function is going to the infinity and the value will be zero. Infinity is a large number and doesn't know the exact value. Any number divided by infinity will be always zero.
Complete step-by-step answer:
The idea of a limit is a basis of all calculus. The limit of a function is defined as let \[f(x)\]be a function defined on an interval that contains\[x = a\]. Then we say that\[\mathop {\lim }\limits_{x \to a} f(x) = L\], if for every \[\varepsilon > 0\]there is some number \[\delta > 0\]such that \[\left| {f(x) - L} \right| < \varepsilon \]whenever\[0 < \left| {x - a} \right| < \delta \].
Here we have to find the value of the limit when x is infinity. The function is trigonometry. Let we define the given function as\[f(x) = \dfrac{{\sin x}}{x}\]. Now we are going to apply the limit to the function \[f(x)\]so we have
\[\mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin x}}{x}\]
The function \[f(x)\] is a trigonometric function. As x approaches \[\infty \] we have to substitute the x as\[\infty \].
When we consider the x value we will consider it as infinity. So by applying limit we have
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } f(x) = \dfrac{{\sin (\infty )}}{\infty }\]
The sine function of infinity is a finite value and the any number divided by infinity is zero so we have
\[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } f(x) = 0\]
Hence the when the x approaches to zero for the function \[\dfrac{{\sin x}}{x}\]
Therefore we have found the limit for the function \[f(x) = \dfrac{{\sin x}}{x}\]
Hence \[\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sin x}}{x} = 0\]
So, the correct answer is “0”.
Note: The function is of trigonometry and it is in form of fraction. The sine function of infinity is some finite value. When the x approaches to infinity, it implies that function is going to the infinity and the value will be zero. Infinity is a large number and doesn't know the exact value. Any number divided by infinity will be always zero.
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