If \(\rm f(x)=\dfrac{\sin x}{x}\) , where x ∈ R, is to be continuous at x = 0, then the value of the function at x = 0
should be 1
Analyzing Function Continuity at a Point
The question asks for the value that the function \(f(x) = \dfrac{\sin x}{x}\) should have at \(x = 0\) to make it continuous at that point. The function is defined as \(f(x)=\dfrac{\sin x}{x}\) for \(x \in \mathbb{R}, x \neq 0\).
Understanding Continuity at a Point
For a function \(f(x)\) to be continuous at a point \(x = a\), three conditions must be met:
- The function must be defined at \(x = a\). That is, \(f(a)\) must exist.
- The limit of the function as \(x\) approaches \(a\) must exist. That is, \(\lim_{x \to a} f(x)\) must exist.
- The value of the function at \(a\) must be equal to the limit of the function as \(x\) approaches \(a\). That is, \(\lim_{x \to a} f(x) = f(a)\).
In this specific problem, we want to ensure continuity at \(x = 0\). The function is given as \(f(x) = \dfrac{\sin x}{x}\) for \(x \neq 0\). Currently, \(f(0)\) is not defined by this expression because the denominator would be zero.
To make the function continuous at \(x = 0\), we need to define or redefine \(f(0)\) such that the third condition of continuity is satisfied. This means we need to set \(f(0)\) equal to the limit of \(f(x)\) as \(x\) approaches 0.
So, we need to evaluate the limit \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \dfrac{\sin x}{x}\).
Evaluating the Limit \(\lim_{x \to 0} \dfrac{\sin x}{x}\)
The limit \(\lim_{x \to 0} \dfrac{\sin x}{x}\) is a fundamental limit in calculus. As \(x\) approaches 0, both \(\sin x\) and \(x\) approach 0, resulting in an indeterminate form of type \(\dfrac{0}{0}\).
There are several ways to evaluate this limit:
Method 1: Using Standard Limit Result
It is a well-known standard limit that:
\(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\)
Method 2: Using L'Hopital's Rule
Since the limit is in the indeterminate form \(\dfrac{0}{0}\), we can apply L'Hopital's Rule. L'Hopital's Rule states that if \(\lim_{x \to a} \dfrac{g(x)}{h(x)}\) is of the form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), then \(\lim_{x \to a} \dfrac{g(x)}{h(x)} = \lim_{x \to a} \dfrac{g'(x)}{h'(x)}\), provided the latter limit exists.
Here, \(g(x) = \sin x\) and \(h(x) = x\). Their derivatives are \(g'(x) = \cos x\) and \(h'(x) = 1\).
So, applying L'Hopital's Rule:
\(\lim_{x \to 0} \dfrac{\sin x}{x} = \lim_{x \to 0} \dfrac{\dfrac{d}{dx}(\sin x)}{\dfrac{d}{dx}(x)} = \lim_{x \to 0} \dfrac{\cos x}{1}\)
Now, substitute \(x = 0\) into the resulting expression:
\(\dfrac{\cos 0}{1} = \dfrac{1}{1} = 1\)
Both methods confirm that \(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\).
Determining the Value for Continuity
For the function \(f(x)\) to be continuous at \(x = 0\), the value of the function at \(x = 0\), i.e., \(f(0)\), must be equal to the limit of the function as \(x\) approaches 0.
We found that \(\lim_{x \to 0} f(x) = \lim_{x \to 0} \dfrac{\sin x}{x} = 1\).
Therefore, for continuity at \(x = 0\), we must define \(f(0)\) such that:
\(f(0) = \lim_{x \to 0} f(x)\)
\(f(0) = 1\)
So, the value of the function at \(x = 0\) should be 1 for it to be continuous at \(x = 0\).
Conclusion
For \(f(x) = \dfrac{\sin x}{x}\) to be continuous at \(x = 0\), the value of the function at \(x = 0\) must be equal to \(\lim_{x \to 0} \dfrac{\sin x}{x}\), which is 1.
The value of the function at \(x = 0\) should be 1.
figure class="table">| Condition for Continuity at \(x=a\) | Application at \(x=0\) |
|---|---|
| \(f(a)\) must be defined | We need to define \(f(0)\). |
| \(\lim_{x \to a} f(x)\) must exist | We evaluate \(\lim_{x \to 0} \dfrac{\sin x}{x}\). This limit exists and is equal to 1. |
| \(\lim_{x \to a} f(x) = f(a)\) | We must set \(f(0) = \lim_{x \to 0} \dfrac{\sin x}{x} = 1\). |
Revision Table: Key Concepts
figure class="table">| Concept | Description |
|---|---|
| Continuity at a point | A function \(f(x)\) is continuous at \(x=a\) if \(f(a)\) is defined, \(\lim_{x \to a} f(x)\) exists, and \(\lim_{x \to a} f(x) = f(a)\). |
| Limit \(\lim_{x \to 0} \dfrac{\sin x}{x}\) | This is a standard limit equal to 1. It can be proven using geometric arguments, Taylor series, or L'Hopital's Rule. |
| Removable Discontinuity | The function \(f(x) = \dfrac{\sin x}{x}\) has a removable discontinuity at \(x=0\) because the limit exists at that point. The discontinuity can be 'removed' by defining or redefining the function value at that point to be equal to the limit. |
Additional Information: Piecewise Functions and Continuity
To make the function \(f(x)\) continuous at \(x = 0\), we can define it as a piecewise function:
\[ g(x) = \begin{cases} \dfrac{\sin x}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]
This new function \(g(x)\) is continuous for all \(x \in \mathbb{R}\). For \(x \neq 0\), \(g(x)\) is a quotient of continuous functions (except where the denominator is zero, which is only at x=0, where this part of the definition doesn't apply). At \(x=0\), we have \(g(0)=1\) (defined), \(\lim_{x \to 0} g(x) = \lim_{x \to 0} \dfrac{\sin x}{x} = 1\) (limit exists), and \(g(0) = \lim_{x \to 0} g(x)\) (\(1=1\)). Thus, all conditions for continuity are met at \(x=0\).
Similar Questions
A function is defined as follows: \[ f(x) = \begin{cases} -\dfrac{x}{\sqrt{x^2}}, & x \ne 0, \\[6pt] 0, & x = 0. \end{cases} \] Which one of the following is correct in respect of the above function?
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Important Questions from Continuity of a function
A function is defined as follows: \[ f(x) = \begin{cases} -\dfrac{x}{\sqrt{x^2}}, & x \ne 0, \\[6pt] 0, & x = 0. \end{cases} \] Which one of the following is correct in respect of the above function?
f(x) = x + |x| is continuous for
Let f : A → R, where A = R\(0) is such that \({\rm{f}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + \left| {\rm{x}} \right|}}{{\rm{x}}}\) . On which one of the following sets is f(x) continuous?
Consider the following statements for f(x) = e -|x| ;
1. The function is continuous at x = 0.
2. The function is differentiable at x = 0.
Which of the above statements is / are correct?
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