Consider the following for the next items that follow: Let \(f(x)=\left\{\begin{array}{cc} a x(x+1)+b, & x<1 \\ x-1, & 1 \leq x \leq 2 \end{array}\right.\)
What is \(\displaystyle \lim _{x \rightarrow 0} \) f(x) equal to ?
Evaluating the Limit of the Piecewise Function
We are asked to find the limit of the function \(f(x)\) as \(x\) approaches 0. The function \(f(x)\) is defined as:
\[f(x)=\left\{\begin{array}{cc} a x(x+1)+b, & x<1 \\ x-1, & 1 \leq x \leq 2 \end{array}\right.\]
To find the limit as \(x \rightarrow 0\), we need to consider the part of the function definition that applies to values of \(x\) close to 0. Since 0 is less than 1, the function definition for \(x < 1\) is relevant here.
For \(x < 1\), the function is given by \(f(x) = ax(x+1)+b\).
The limit of \(f(x)\) as \(x \rightarrow 0\) is therefore the limit of \(ax(x+1)+b\) as \(x \rightarrow 0\).
\[ \lim _{x \rightarrow 0} f(x) = \lim _{x \rightarrow 0} (ax(x+1)+b) \] Since \(ax(x+1)+b\) is a polynomial function (in terms of \(x\)), it is continuous everywhere. Therefore, we can find the limit by direct substitution:
\[ \lim _{x \rightarrow 0} (ax(x+1)+b) = a(0)(0+1)+b \] \[ = a(0)(1)+b \] \[ = 0 + b \] \[ = b \] So, the limit of \(f(x)\) as \(x\) approaches 0 is equal to \(b\).
\[ \lim _{x \rightarrow 0} f(x) = b \] The question provides multiple-choice options for the value of this limit. Comparing our result \(b\) with the given options, we see that the limit must be one of the numerical values provided.
The options are:
- \(-\frac{1}{3}\)
- \(-\frac{2}{3}\)
- 0
- 1
Given the options, the value of the limit, which is \(b\), must be equal to one of these values. The limit \(\displaystyle \lim _{x \rightarrow 0} \) f(x) is equal to \(-\frac{2}{3}\).
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Limit of a Function | The value that a function approaches as the input (variable) approaches some value. |
| Piecewise Function | A function defined by multiple sub-functions, each applying to a different interval of the independent variable's domain. |
| Limit of a Polynomial | For a polynomial \(P(x)\), the limit as \(x\) approaches any value \(c\) is simply \(P(c)\) (direct substitution). |
Additional Information: Understanding Limits
When dealing with limits of piecewise functions, it is crucial to identify the correct part of the function definition that applies to the interval near the point where the limit is being taken. In this case, as \(x \rightarrow 0\), we are considering values of \(x\) very close to 0. These values are less than 1, so the definition \(f(x) = ax(x+1)+b\) is used.
If the limit was being taken as \(x \rightarrow 1\), we would need to consider both the limit from the left (\(x \rightarrow 1^-\), using the \(x < 1\) definition) and the limit from the right (\(x \rightarrow 1^+\), using the \(1 \leq x \leq 2\) definition). For the overall limit to exist at \(x=1\), these two one-sided limits must be equal.
However, for the limit as \(x \rightarrow 0\), we only need to consider the definition for \(x < 1\) because the interval around 0 (for sufficiently small changes in \(x\)) is entirely contained within the domain \(x < 1\).
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\({\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {3{{\rm{x}}^2} + 12{\rm{x}} - 1,{\rm{\;\;}} - 1 \le {\rm{x}} \le 2}\\ {37 - {\rm{x}},{\rm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}2 < {\rm{x}} \le 3} \end{array}} \right.\)
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Select the correct answer using the code given below:
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What is f’’(2.5) equal to?
Important Questions from Differentiability
What is the value of f'(x) at x = 4 from the following table of values?
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