The function \(f(x)=\left\{\begin{matrix}\dfrac{|x|}{3x^2-5x},\ x\ne0 \\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0\end{matrix}\right.\)
is not continuous at x = 0, because
Understanding Function Continuity at a Point
To determine if a function \(f(x)\) is continuous at a specific point, say \(x=c\), we need to check three conditions:
- The function must be defined at \(c\), i.e., \(f(c)\) exists.
- The limit of the function as \(x\) approaches \(c\) must exist, i.e., \(\displaystyle\lim_{x\rightarrow c} f(x)\) exists.
- The limit of the function as \(x\) approaches \(c\) must be equal to the function's value at \(c\), i.e., \(\displaystyle\lim_{x\rightarrow c} f(x) = f(c)\).
If any of these conditions are not met, the function is not continuous at \(x=c\).
Analyzing the Given Function \(f(x)\) at x = 0
The function is given as:
\[f(x)=\left\{\begin{matrix}\dfrac{|x|}{3x^2-5x},\ x\ne0 \\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0\end{matrix}\right.\]
We need to check its continuity at \(x = 0\).
Step 1: Check if \(f(0)\) exists
From the definition of the function, at \(x = 0\), \(f(0) = 0\). So, \(f(0)\) exists.
Step 2: Check if the limit \(\displaystyle\lim_{x\rightarrow0}f(x)\) exists
For the limit \(\displaystyle\lim_{x\rightarrow0}f(x)\) to exist, the left-hand limit (LHL) and the right-hand limit (RHL) at \(x=0\) must be equal.
Calculating the Left-Hand Limit (LHL) at x = 0
The LHL is \(\displaystyle\lim_{x\rightarrow0^-}f(x)\). When \(x\) approaches 0 from the left side, \(x < 0\). For \(x < 0\), the definition of the function is \(f(x) = \dfrac{|x|}{3x^2-5x}\). Since \(x < 0\), \(|x| = -x\).
\[\lim_{x\rightarrow0^-}f(x) = \lim_{x\rightarrow0^-}\dfrac{|x|}{3x^2-5x} = \lim_{x\rightarrow0^-}\dfrac{-x}{3x^2-5x}\]
We can factor out \(x\) from the denominator:
\[\lim_{x\rightarrow0^-}\dfrac{-x}{x(3x-5)}\]
For \(x \ne 0\), we can cancel out \(x\):
\[\lim_{x\rightarrow0^-}\dfrac{-1}{3x-5}\]
Now, substitute \(x=0\):
\[\dfrac{-1}{3(0)-5} = \dfrac{-1}{-5} = \dfrac{1}{5}\]
So, the LHL is \(\dfrac{1}{5}\).
Calculating the Right-Hand Limit (RHL) at x = 0
The RHL is \(\displaystyle\lim_{x\rightarrow0^+}f(x)\). When \(x\) approaches 0 from the right side, \(x > 0\). For \(x > 0\), the definition of the function is \(f(x) = \dfrac{|x|}{3x^2-5x}\). Since \(x > 0\), \(|x| = x\).
\[\lim_{x\rightarrow0^+}f(x) = \lim_{x\rightarrow0^+}\dfrac{|x|}{3x^2-5x} = \lim_{x\rightarrow0^+}\dfrac{x}{3x^2-5x}\]
Factor out \(x\) from the denominator:
\[\lim_{x\rightarrow0^+}\dfrac{x}{x(3x-5)}\]
For \(x \ne 0\), we can cancel out \(x\):
\[\lim_{x\rightarrow0^+}\dfrac{1}{3x-5}\]
Now, substitute \(x=0\):
\[\dfrac{1}{3(0)-5} = \dfrac{1}{-5} = -\dfrac{1}{5}\]
So, the RHL is \(-\dfrac{1}{5}\).
Comparing LHL and RHL
We found that LHL \(=\) \(\dfrac{1}{5}\) and RHL \(=\) \(-\dfrac{1}{5}\). Since LHL \(\ne\) RHL, the limit \(\displaystyle\lim_{x\rightarrow0}f(x)\) does not exist.
Step 3: Check if \(\displaystyle\lim_{x\rightarrow0}f(x) = f(0)\)
Since the limit \(\displaystyle\lim_{x\rightarrow0}f(x)\) does not exist (from Step 2), the third condition for continuity cannot be met.
Conclusion on Continuity at x = 0
Because the limit of the function as \(x\) approaches 0, \(\displaystyle\lim_{x\rightarrow0}f(x)\), does not exist, the function \(f(x)\) is not continuous at \(x=0\). This is the primary reason for the discontinuity.
Analyzing the Given Options
Let's evaluate the given options based on our findings:
- Option 1: \(\displaystyle\lim_{x\rightarrow0}f(x)\ne{f(0)}\) - This condition implies the limit exists but is not equal to the function value. Our analysis showed the limit does not exist, so this isn't the direct reason, although it's a consequence.
- Option 2: \(\displaystyle\lim_{x\rightarrow0^-}f(x)\) does not exist - We calculated the LHL to be \(\dfrac{1}{5}\), which exists. So this option is incorrect.
- Option 3: \(\displaystyle\lim_{x\rightarrow0}f(x)\) does not exist - Our calculation showed that LHL \(\ne\) RHL, therefore the overall limit \(\displaystyle\lim_{x\rightarrow0}f(x)\) does not exist. This is correct.
- Option 4: \(\displaystyle\lim_{x\rightarrow0^+}f(x)\) does not exist - We calculated the RHL to be \(-\dfrac{1}{5}\), which exists. So this option is incorrect.
The main reason the function is not continuous at \(x=0\) is that the overall limit at \(x=0\) fails to exist because the left-hand limit and the right-hand limit are not equal.
| Continuity Condition | Status at x=0 | Explanation |
|---|---|---|
| \(f(0)\) exists? | Yes | \(f(0) = 0\) (given) |
| \(\displaystyle\lim_{x\rightarrow0}f(x)\) exists? | No | LHL (\(\dfrac{1}{5}\)) \(\ne\) RHL (\(-\dfrac{1}{5}\)) |
| \(\displaystyle\lim_{x\rightarrow0}f(x) = f(0)\)? | No | Limit does not exist, so equality cannot hold |
Revision Table: Key Concepts for Function Continuity
| Concept | Definition/Meaning | Relevance to Continuity |
|---|---|---|
| Continuity at \(x=c\) | Function is defined at \(c\), limit exists at \(c\), and limit equals \(f(c)\). | The fundamental property being checked. |
| Limit \(\displaystyle\lim_{x\rightarrow c}f(x)\) | The value the function approaches as \(x\) gets arbitrarily close to \(c\) (from both sides), without necessarily being equal to \(c\). | Must exist for continuity. |
| Left-Hand Limit (\(\displaystyle\lim_{x\rightarrow c^-}f(x)\)) | The value \(f(x)\) approaches as \(x\) approaches \(c\) from values less than \(c\). | Must equal RHL for the overall limit to exist. |
| Right-Hand Limit (\(\displaystyle\lim_{x\rightarrow c^+}f(x)\)) | The value \(f(x)\) approaches as \(x\) approaches \(c\) from values greater than \(c\). | Must equal LHL for the overall limit to exist. |
| Discontinuity | A point where the function is not continuous. | Can occur if \(f(c)\) doesn't exist, the limit doesn't exist, or the limit doesn't equal \(f(c)\). |
Additional Information: Types of Discontinuity
When a function is not continuous at a point \(x=c\), it has a discontinuity. There are different types of discontinuity:
- Removable Discontinuity: The limit \(\displaystyle\lim_{x\rightarrow c} f(x)\) exists, but either \(f(c)\) does not exist or \(\displaystyle\lim_{x\rightarrow c} f(x) \ne f(c)\). These are often "holes" in the graph.
- Jump Discontinuity: The left-hand limit and the right-hand limit both exist, but they are not equal (\(\displaystyle\lim_{x\rightarrow c^-} f(x) \ne \lim_{x\rightarrow c^+} f(x)\)). This causes a "jump" in the graph. In our case, the function \(f(x)\) at \(x=0\) exhibits a jump discontinuity because LHL \(\ne\) RHL.
- Infinite Discontinuity: At least one of the one-sided limits (\(\displaystyle\lim_{x\rightarrow c^-} f(x)\) or \(\displaystyle\lim_{x\rightarrow c^+} f(x)\)) is either \(+\infty\) or \(-\infty\). These typically occur at vertical asymptotes.
- Essential Discontinuity: Neither a removable nor a jump discontinuity. Often occurs when one or both of the one-sided limits do not exist in any finite sense (e.g., \(\displaystyle\lim_{x\rightarrow 0} \sin(1/x)\)).
Our function \(f(x)\) at \(x=0\) has a jump discontinuity because the LHL and RHL are finite but unequal.
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?
If the function \(\rm f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {a + bx,\;\;}&{x < 1}\\ {5,}&{x = 1}\\ {b - ax,}&{x > 1} \end{array}} \right.\) is continuous, then what is the value of (a + b)?
