Limits Involving Algebraic Functions & Piecewise Definitions
In this reading, we connect our understanding of limits to a very important practical topic: limits of piecewise functions in calculus. Up to now, most examples involved a single formula valid for all real numbers, such as polynomials, rational functions, exponentials, or trigonometric expressions. In real situations, however, functions are often defined by different formulas on different intervals. That is where limits of piecewise functions in calculus become essential.
Our goals are to:
- Review basic limit behaviour for algebraic functions.
- Understand what a piecewise function is and how to compute limits of piecewise functions in calculus.
- Study jump discontinuities, removable discontinuities, and examples where limits exist or do not exist.
1. Quick recall: limits of algebraic functions
Before focusing on limits of piecewise functions in calculus, we briefly recall what happens for simple algebraic functions:
- Polynomials: For a polynomial ##p(x)##, the limit at any real number ##a## is just ##p(a)##:
###\lim_{x \to a} p(x) = p(a).###
Polynomials are continuous everywhere. - Rational functions: For a rational function ##\dfrac{p(x)}{q(x)}##, where ##p## and ##q## are polynomials,
###\lim_{x \to a} \frac{p(x)}{q(x)} = \frac{p(a)}{q(a)}###
whenever ##q(a) \neq 0##. - If ##q(a) = 0##, then we may obtain forms like ##\dfrac{0}{0}## or unbounded behaviour. In such cases, factorisation and algebraic simplification are required.
For “single-formula” functions, many limits are straightforward. The main challenge in this reading appears when we turn to limits of piecewise functions in calculus, where the definition itself changes across intervals.
2. What is a piecewise function?
A piecewise function is defined by different formulas on different parts of its domain. For example,
###f(x) =
\begin{cases}
x^2, & x < 1, \\
2x – 1, & x \ge 1
\end{cases}###
For ##x < 1##, the function follows ##x^2##; for ##x \ge 1##, it follows ##2x – 1##. When we talk about limits of piecewise functions in calculus, the behaviour at the boundary point (here ##x = 1##) becomes especially important.
For interior points of one branch, such as ##x = -2## or ##x = 3## in the example, limits behave just as they do for ordinary algebraic functions. The real work lies at the joining points where the formula switches.
3. Left-hand and right-hand limits
For limits of piecewise functions in calculus, the left-hand limit (LHL) and right-hand limit (RHL) at a point ##x = a## play a crucial role:
- Left-hand limit (LHL):
###\lim_{x \to a^-} f(x)###
describes the value approached as ##x## approaches ##a## from the left (values less than ##a##). - Right-hand limit (RHL):
###\lim_{x \to a^+} f(x)###
describes the value approached as ##x## approaches ##a## from the right (values greater than ##a##).
For a limit to exist at ##x = a##, both limits must exist and be equal:
###\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L.###
Only then do we say that ###\lim_{x \to a} f(x) = L###.
4. Worked examples: basic piecewise limits
Example 1
Let
###f(x) =
\begin{cases}
x^2, & x < 1, \\
2x – 1, & x \ge 1.
\end{cases}###
Find ###\displaystyle \lim_{x \to 1} f(x).###
Solution.
For ##x < 1##, we use ##f(x) = x^2##; for ##x \ge 1##, we use ##f(x) = 2x – 1##.
Left-hand limit:
###\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x^2 = 1^2 = 1.###
Right-hand limit:
###\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (2x – 1) = 2 \cdot 1 – 1 = 1.###
Since LHL = RHL = 1, we conclude
###\lim_{x \to 1} f(x) = 1.###
Example 2
Define
###g(x) =
\begin{cases}
2, & x < 0, \\
5, & x \ge 0.
\end{cases}###
Investigate ###\displaystyle \lim_{x \to 0} g(x).###
Solution.
Left-hand limit:
For ##x < 0##, ##g(x) = 2##, so
###\lim_{x \to 0^-} g(x) = 2.###
Right-hand limit:
For ##x \ge 0##, ##g(x) = 5##, so
###\lim_{x \to 0^+} g(x) = 5.###
Here LHL = 2 and RHL = 5. They are not equal, so
###\lim_{x \to 0} g(x) \text{ does not exist}.###
This is a jump discontinuity, a common feature of limits of piecewise functions in calculus.
Example 3
Let
###h(x) =
\begin{cases}
x + 2, & x < 1, \\
3 – x, & x \ge 1.
\end{cases}###
Find ###\displaystyle \lim_{x \to 1} h(x)### and identify the type of discontinuity (if any).
Solution.
Left-hand limit:
###\lim_{x \to 1^-} h(x) = \lim_{x \to 1^-} (x + 2) = 3.###
Right-hand limit:
###\lim_{x \to 1^+} h(x) = \lim_{x \to 1^+} (3 – x) = 2.###
Since LHL = 3 and RHL = 2, the limit does not exist. Both one-sided limits are finite but unequal, so the function has a jump discontinuity at ##x = 1##.
5. Removable discontinuities
Sometimes the limit exists, but the function value at that point is different or undefined. Such a point is called a removable discontinuity, because redefining the function at that point can make it continuous.
Example 4
Consider
###f(x) =
\begin{cases}
x^2, & x \neq 2, \\
5, & x = 2.
\end{cases}###
Find ###\displaystyle \lim_{x \to 2} f(x)### and comment on continuity at ##x = 2##.
Solution.
For all ##x \ne 2##, the formula is ##x^2##, so
###\lim_{x \to 2} f(x) = \lim_{x \to 2} x^2 = 4.###
However, ##f(2) = 5##. The limit exists and equals 4, but the function value at 2 is 5. Thus the function is not continuous at 2, and the discontinuity is removable: if we redefine ##f(2) = 4##, the function becomes continuous.
6. Absolute value functions as piecewise definitions
Absolute value functions are natural examples of limits of piecewise functions in calculus. For instance,
###|x| =
\begin{cases}
-x, & x < 0, \\
x, & x \ge 0.
\end{cases}###
Example 5
Consider
###f(x) = \frac{|x|}{x}, \quad x \neq 0.###
Examine ###\displaystyle \lim_{x \to 0} \frac{|x|}{x}.###
Solution.
Write ##f(x)## explicitly:
###\frac{|x|}{x} =
\begin{cases}
\frac{-x}{x} = -1, & x < 0, \\
\frac{x}{x} = 1, & x > 0.
\end{cases}###
Then
- LHL: ###\displaystyle \lim_{x \to 0^-} \frac{|x|}{x} = -1.###
- RHL: ###\displaystyle \lim_{x \to 0^+} \frac{|x|}{x} = 1.###
Since LHL ≠ RHL, the two-sided limit does not exist. This is another jump discontinuity example.
7. More worked examples of limits of piecewise functions
Example 6
Let
###f(x) =
\begin{cases}
x^2 – 4, & x \neq 2, \\
0, & x = 2.
\end{cases}###
Find ###\displaystyle \lim_{x \to 2} f(x)### and determine whether ##f## is continuous at ##x = 2##.
Solution.
For ##x \ne 2##, ##f(x) = x^2 – 4##, so
###\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 – 4) = 4 – 4 = 0.###
We also have ##f(2) = 0##. Thus
###\lim_{x \to 2} f(x) = f(2) = 0.###
So ##f## is continuous at ##x = 2##. In this example, although the function is defined piecewise, there is no discontinuity at the joining point.
Example 7
Let
###f(x) =
\begin{cases}
x – 1, & x < 2, \\
x^2 – 4, & x \ge 2.
\end{cases}###
Find ###\displaystyle \lim_{x \to 2} f(x)### and comment on continuity at ##x = 2##.
Solution.
LHL:
###\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x – 1) = 1.###
RHL:
###\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 – 4) = 0.###
LHL ≠ RHL, so ###\lim_{x \to 2} f(x)### does not exist and the function has a jump discontinuity at ##x = 2##.
Example 8
Define
###f(x) =
\begin{cases}
x^2, & x < 0, \\
0, & x = 0, \\
x, & x > 0.
\end{cases}###
Find the limits ###\displaystyle \lim_{x \to 0^-} f(x), \lim_{x \to 0^+} f(x), \lim_{x \to 0} f(x)### and determine continuity at 0.
Solution.
LHL:
###\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0.###
RHL:
###\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0.###
Thus ###\lim_{x \to 0} f(x) = 0###. Since ##f(0) = 0##, the function is continuous at 0.
Example 9
Let
###f(x) =
\begin{cases}
k x + 1, & x < 1, \\
x^2, & x \ge 1.
\end{cases}###
Find the value of ##k## for which ##\displaystyle \lim_{x \to 1} f(x)## exists and compute the limit for that value of ##k##.
Solution.
For the limit to exist, LHL must equal RHL at ##x = 1##.
LHL:
###\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (kx + 1) = k \cdot 1 + 1 = k + 1.###
RHL:
###\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1.###
Equating:
###k + 1 = 1 \Rightarrow k = 0.###
For ##k = 0##, both LHL and RHL equal 1, and therefore
###\lim_{x \to 1} f(x) = 1.###
Example 10
Construct a piecewise function with a removable discontinuity at ##x = 1## and a jump discontinuity at ##x = 2##, and compute the limits at those points.
Solution.
Define
###f(x) =
\begin{cases}
x, & x \neq 1, x < 2, \\
3, & x = 1, \\
x + 1, & x \ge 2.
\end{cases}###
At ##x = 1##:
- LHL: ###\displaystyle \lim_{x \to 1^-} f(x) = 1.###
- RHL: ###\displaystyle \lim_{x \to 1^+} f(x) = 1.###
Thus ###\lim_{x \to 1} f(x) = 1### but ##f(1) = 3##, so there is a removable discontinuity at 1.
At ##x = 2##:
- LHL: for ##x < 2##, ##f(x) = x##, so ###\lim_{x \to 2^-} f(x) = 2.###
- RHL: for ##x \ge 2##, ##f(x) = x + 1##, so ###\lim_{x \to 2^+} f(x) = 3.###
LHL ≠ RHL, so ###\lim_{x \to 2} f(x)### does not exist and there is a jump discontinuity at 2.
8. Summary and viewpoint
In the context of limits of piecewise functions in calculus, at a point ##x = a## we typically see one of these patterns:
- Both LHL and RHL exist and are equal: the limit exists and equals this common value.
- LHL and RHL exist but differ: the two-sided limit does not exist; this is often a jump discontinuity.
- At least one side diverges or oscillates: the limit does not exist (possibly an infinite limit).
With this systematic perspective, we can analyse complicated piecewise definitions step by step.
9. References for further reading
- The article on functions and continuity at Wikipedia discusses piecewise definitions and types of discontinuities with graphs.
- Introductory notes from organisations such as the American Mathematical Society (AMS) and the National Institute of Standards and Technology (NIST) provide additional context on limits and continuity in analysis.
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