Consider the following in respect of the function \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left\{ {\begin{array}{} {2 + {\rm{x}},{\rm{\;\;x}} \ge 0}\\ {2 - {\rm{x}},{\rm{\;\;x}} < 0} \end{array}} \right.\) 1. \(\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\) Does not exists. 2. f(x) is differentiable at x = 0 3. f(x) is continuous at x = 0 Which of the above statement is/are correct?
3 only
Analyzing Piecewise Function Properties
The question asks us to consider a piecewise function \({\rm{f}}\left( {\rm{x}} \right)\) and evaluate three statements regarding its limit, differentiability, and continuity.
The function is defined as:
\({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left\{ {\begin{array}{} {2 + {\rm{x}},{\rm{\;\;x}} \ge 0}\\ {2 - {\rm{x}},{\rm{\;\;x}} < 0} \end{array}} \right.\)
Let's analyze each statement carefully.
Statement 1: \(\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\) Does not exist.
To evaluate the limit of \({\rm{f}}\left( {\rm{x}} \right)\) as \({\rm{x}}\) approaches 1, we need to look at the function definition for values of \({\rm{x}}\) near 1. Since \({\rm{x}} = 1\) is greater than or equal to 0, the definition of the function for \({\rm{x}} \ge 0\) applies when evaluating the limit as \({\rm{x}} \to 1\).
For \({\rm{x}} \ge 0\), \({\rm{f}}\left( {\rm{x}} \right) = 2 + {\rm{x}}\). Therefore, the limit is:
\(\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to 1} \left( {2 + {\rm{x}}} \right)\)
Substituting \({\rm{x}} = 1\) into the expression, we get:
\(\mathop {\lim }\limits_{{\rm{x}} \to 1} \left( {2 + {\rm{x}}} \right) = 2 + 1 = 3\)
The limit of \({\rm{f}}\left( {\rm{x}} \right)\) as \({\rm{x}} \to 1\) exists and is equal to 3. Thus, Statement 1 is incorrect.
Statement 2: f(x) is differentiable at x = 0
For a function to be differentiable at a point \({\rm{x}} = {\rm{a}}\), its left-hand derivative and right-hand derivative at that point must exist and be equal. The derivative at \({\rm{x}} = {\rm{a}}\) is defined as:
\({\rm{f}}'\left( {\rm{a}} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{\rm{f}}\left( {{\rm{a}} + h} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{h}\)
We need to check differentiability at \({\rm{x}} = 0\). First, let's find \({\rm{f}}\left( 0 \right)\). Since \(0 \ge 0\), we use the definition \({\rm{f}}\left( {\rm{x}} \right) = 2 + {\rm{x}}\).
\({\rm{f}}\left( 0 \right) = 2 + 0 = 2\)
Now, let's calculate the left-hand derivative (LHD) at \({\rm{x}} = 0\):
\(\text{LHD} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{{\rm{f}}\left( {0 + h} \right) - {\rm{f}}\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{{\rm{f}}\left( h \right) - 2}}{h}\)
For \(h < 0\), \({\rm{f}}\left( h \right) = 2 - h\). So,
\(\text{LHD} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{\left( {2 - h} \right) - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \frac{{ - h}}{h} = \mathop {\lim }\limits_{h \to {0^ - }} \left( { - 1} \right) = - 1\)
Next, let's calculate the right-hand derivative (RHD) at \({\rm{x}} = 0\):
\(\text{RHD} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{{\rm{f}}\left( {0 + h} \right) - {\rm{f}}\left( 0 \right)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{{\rm{f}}\left( h \right) - 2}}{h}\)
For \(h > 0\), \({\rm{f}}\left( h \right) = 2 + h\). So,
\(\text{RHD} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{\left( {2 + h} \right) - 2}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{h}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 1 = 1\)
Since the LHD (\(-1\)) is not equal to the RHD (\(1\)), the function \({\rm{f}}\left( {\rm{x}} \right)\) is not differentiable at \({\rm{x}} = 0\). Thus, Statement 2 is incorrect.
Statement 3: f(x) is continuous at x = 0
For a function to be continuous at a point \({\rm{x}} = {\rm{a}}\), three conditions must be met:
- \({\rm{f}}\left( {\rm{a}} \right)\) must be defined.
- \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\) must exist.
- \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\).
We need to check continuity at \({\rm{x}} = 0\).
Condition 1: \({\rm{f}}\left( 0 \right)\) is defined, as calculated before: \({\rm{f}}\left( 0 \right) = 2\).
Condition 2: \(\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right)\) must exist. For the limit to exist at \({\rm{x}} = 0\), the left-hand limit (LHL) and the right-hand limit (RHL) must exist and be equal.
LHL at \({\rm{x}} = 0\): \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} {\rm{f}}\left( {\rm{x}} \right)\). For \({\rm{x}} < 0\), \({\rm{f}}\left( {\rm{x}} \right) = 2 - {\rm{x}}\).
\(\text{LHL} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ - }} \left( {2 - {\rm{x}}} \right) = 2 - 0 = 2\)
RHL at \({\rm{x}} = 0\): \(\mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} {\rm{f}}\left( {\rm{x}} \right)\). For \({\rm{x}} > 0\), \({\rm{f}}\left( {\rm{x}} \right) = 2 + {\rm{x}}\).
\(\text{RHL} = \mathop {\lim }\limits_{{\rm{x}} \to {0^ + }} \left( {2 + {\rm{x}}} \right) = 2 + 0 = 2\)
Since LHL = RHL = 2, the limit \(\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right)\) exists and is equal to 2.
Condition 3: \(\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( 0 \right)\). We have \(\mathop {\lim }\limits_{{\rm{x}} \to 0} {\rm{f}}\left( {\rm{x}} \right) = 2\) and \({\rm{f}}\left( 0 \right) = 2\). Since \(2 = 2\), this condition is met.
All three conditions for continuity at \({\rm{x}} = 0\) are satisfied. Therefore, the function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}} = 0\). Thus, Statement 3 is correct.
Summary of Analysis
| Statement | Claim | Result | Correctness |
|---|---|---|---|
| 1 | \(\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\) Does not exist | Limit is 3 | Incorrect |
| 2 | f(x) is differentiable at x = 0 | LHD \(\ne\) RHD (-1 \(\ne\) 1) | Incorrect |
| 3 | f(x) is continuous at x = 0 | LHL = RHL = f(0) = 2 | Correct |
Based on the analysis, only Statement 3 is correct.
Conclusion on Function Properties
We have evaluated the given piecewise function \({\rm{f}}\left( {\rm{x}} \right)\) for its limit at \({\rm{x}}=1\), differentiability at \({\rm{x}}=0\), and continuity at \({\rm{x}}=0\).
- The limit \(\mathop {\lim }\limits_{{\rm{x}} \to 1} {\rm{f}}\left( {\rm{x}} \right)\) exists and equals 3.
- The function \({\rm{f}}\left( {\rm{x}} \right)\) is not differentiable at \({\rm{x}}=0\) because the left-hand derivative and right-hand derivative are not equal. This point \({\rm{x}}=0\) is a sharp corner or cusp in the graph of the function (it's related to the absolute value function \(|{\rm{x}}|\)).
- The function \({\rm{f}}\left( {\rm{x}} \right)\) is continuous at \({\rm{x}}=0\) because the limit exists at \({\rm{x}}=0\) and is equal to the function value at \({\rm{x}}=0\).
Therefore, only the third statement is correct.
Revision Table: Function Analysis Concepts
| Concept | Definition (at x=a) | How to Check for Piecewise Functions |
|---|---|---|
| Limit (\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\)) | Value the function approaches as x gets arbitrarily close to 'a' (from both sides). | Evaluate LHL (\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}^ - } {\rm{f}}\left( {\rm{x}} \right)\)) and RHL (\(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}^ + } {\rm{f}}\left( {\rm{x}} \right)\)). Limit exists if LHL = RHL. For points not at the boundary, use the single definition. |
| Continuity (at x=a) | The graph can be drawn without lifting the pen. | Check if 1) \({\rm{f}}\left( {\rm{a}} \right)\) is defined, 2) \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right)\) exists, and 3) \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{a}} \right)\). Condition 2 involves checking LHL and RHL equality. |
| Differentiability (at x=a) | The function has a well-defined tangent line at 'a'. Smoothness of the graph. | Check if LHD (\(\mathop {\lim }\limits_{h \to {0^ - }} \frac{{{\rm{f}}\left( {{\rm{a}} + h} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{h}\)) equals RHD (\(\mathop {\lim }\limits_{h \to {0^ + }} \frac{{{\rm{f}}\left( {{\rm{a}} + h} \right) - {\rm{f}}\left( {\rm{a}} \right)}}{h}\)). If a function is differentiable at 'a', it must also be continuous at 'a'. The converse is not always true. |
Additional Information: Relationship between Concepts
It's important to understand the relationship between continuity and differentiability. If a function is differentiable at a point, it must be continuous at that point. However, the reverse is not true. A function can be continuous at a point but not differentiable there, as demonstrated by the function in this problem at \({\rm{x}}=0\). The graph of \({\rm{f}}\left( {\rm{x}} \right)\) at \({\rm{x}}=0\) has a sharp point, indicating continuity but not differentiability.
The function \({\rm{f}}\left( {\rm{x}} \right)\) in this problem is essentially \({\rm{f}}\left( {\rm{x}} \right) = 2 + |{\rm{x}}|\). The absolute value function \(|{\rm{x}}|\) is known to be continuous everywhere but not differentiable at \({\rm{x}} = 0\). Adding a constant (2) shifts the graph vertically but does not change its continuity or differentiability properties at specific points.
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