Why the Side Matters
An absolute value can change the expression being evaluated when its input changes sign. Near a target input, that sign may be different to the left and to the right. Consequently, an expression involving absolute values can have different one-sided limits, even when the algebra inside the absolute values looks simple.
The central task is to determine the sign of each enclosed expression on the side being considered. After that, replace each absolute value with the appropriate expression and evaluate the resulting one-sided limit using the usual limit laws.
The Absolute-Value Rule
For any real number \(u\), absolute value is defined by
In a limit as \(x\) approaches \(a\), the input \(u\) may be an expression depending on \(x\). The value at \(x=a\) alone is not enough to choose a branch: what matters is the sign of \(u(x)\) for \(x\) close to \(a\) on the specified side.
For example, \(x-a\) is negative when \(x<a\) and positive when \(x>a\). Thus the same absolute value has two different local formulas:
These identities hold on the indicated intervals, not just at individual points. That makes them useful for evaluating one-sided limits.
A Sign-Resolution Method
When several absolute values occur, handle their inputs one at a time. A sign determination for one input does not automatically settle the sign of another. Factorization is often useful: it separates factors whose signs can be checked individually.
Decide whether \(x\) approaches \(a\) from below or above. Keep that side fixed throughout the calculation.
Factor or otherwise analyze each enclosed expression to determine its sign close to \(a\) on that side.
Use the input itself when it is nonnegative, and its negative when it is negative. Do this separately for every absolute value.
Once the expression is correct on the chosen interval, apply the relevant one-sided limit laws and simplify.
A two-sided limit exists only if both one-sided limits exist and are equal.
Why the Sign Can Be Fixed Near the Target
A basic reason the method works is that a continuous expression which is nonzero at the target keeps its sign sufficiently close to that target. In particular, if \(g\) is continuous at \(a\) and \(g(a)>0\), then \(g(x)>0\) for all \(x\) sufficiently close to \(a\). If \(g(a)<0\), then \(g(x)<0\) sufficiently close to \(a\). This follows from continuity: choose a neighborhood in which \(|g(x)-g(a)|<|g(a)|\). In that neighborhood, \(g(x)\) cannot reach zero or change sign.
If \(g(a)=0\), its sign may differ on the two sides. A common case is a factorization
where \(m\) is a positive integer and \(h\) is continuous with \(h(a)\ne 0\). Near \(a\), \(h\) keeps the sign of \(h(a)\). The factor \((x-a)^m\) is positive on both sides when \(m\) is even; when \(m\) is odd, it is negative to the left and positive to the right. This explains why checking the side and the factorization resolves the absolute value.
Worked Examples
Worked Example: A Basic Absolute Value at Its Zero
Evaluate the left- and right-hand limits of \(f(x)=\dfrac{|x-2|}{x-2}\) as \(x\to 2\).
From the left: If \(x<2\), then \(x-2<0\), so \(|x-2|=-(x-2)\). For every \(x<2\) close to 2,
Therefore, \(\displaystyle\lim_{x\to 2^-}\frac{|x-2|}{x-2}=-1\).
From the right: If \(x>2\), then \(x-2>0\), so \(|x-2|=x-2\). For every \(x>2\) close to 2,
Therefore, \(\displaystyle\lim_{x\to 2^+}\frac{|x-2|}{x-2}=1\). The two one-sided limits are unequal, so the two-sided limit does not exist. As a check, the original quotient is exactly \(-1\) on the left interval and exactly \(1\) on the right interval, confirming both results.
Worked Example: Factoring to Determine the Sign
Evaluate \(\displaystyle\lim_{x\to 1^-}\frac{|x^2-1|}{x-1}\) and \(\displaystyle\lim_{x\to 1^+}\frac{|x^2-1|}{x-1}\).
Factor the expression inside the absolute value:
For \(x\) sufficiently close to \(1\), the factor \(x+1\) is positive. Thus the sign of \(x^2-1\) is determined by \(x-1\).
From the left: When \(x<1\), \(x-1<0\) and \(x+1>0\), so \(x^2-1<0\). Hence \(|x^2-1|=-(x^2-1)\), and for \(x<1\) near 1,
Therefore, the left-hand limit is \(-(1+1)=-2\).
From the right: When \(x>1\), both \(x-1\) and \(x+1\) are positive, so \(x^2-1>0\). Thus \(|x^2-1|=x^2-1\), and
The right-hand limit is \(1+1=2\). As a check, the sign-specific simplified expressions are \(-(x+1)\) on the left and \(x+1\) on the right; their respective limits are \(-2\) and \(2\), as calculated. Since the one-sided limits differ, the two-sided limit does not exist.
Worked Example: Resolving Two Absolute Values Separately
Evaluate the one-sided limits of \(f(x)=\dfrac{|x-1|}{(x-1)|x+2|}\) as \(x\to 1\).
There are two absolute values, so determine both signs. Near \(x=1\), the sign of \(x-1\) depends on the side. Meanwhile, \(x+2>0\) on both sides sufficiently close to 1, so \(|x+2|=x+2\) on both sides.
From the left: For \(x<1\), \(|x-1|=-(x-1)\). Therefore,
Taking the left-hand limit gives \(-\dfrac{1}{1+2}=-\dfrac{1}{3}\).
From the right: For \(x>1\), \(|x-1|=x-1\), while \(|x+2|=x+2\) still holds. Thus
The right-hand limit is \(\dfrac{1}{1+2}=\dfrac{1}{3}\). As a check, the original function equals \(-1/(x+2)\) on the left and \(1/(x+2)\) on the right, giving the stated limits. The absolute value \(|x+2|\) does not change its formula here; it must still be checked rather than assumed to behave like \(|x-1|\).
A Common Sign Error
A frequent mistake is to replace \(|x-a|\) by \(x-a\) on both sides of \(a\). For instance, an incorrect calculation might claim
The replacement \(|x-2|=x-2\) is valid only when \(x\geq 2\), not when \(x\to 2^-\). On the left, \(x-2<0\), so the correct replacement is \(|x-2|=-(x-2)\). Therefore the quotient is \(-1\), and the left-hand limit is \(-1\). The correction is not to memorize a preferred sign, but to establish the sign on the requested side before removing the absolute value.
When the Two-Sided Limit Exists
After evaluating the one-sided limits, compare them. If both exist as the same finite number, the two-sided limit exists and equals that number. If they differ, the two-sided limit does not exist. The first worked example shows a common absolute-value pattern: the function has one constant value on one side and a different constant value on the other.
It is useful to keep the calculation organized by side, especially when multiple absolute values occur. A small sign table can prevent a replacement on one side from accidentally being carried over to the other:
| Expression | Sign or replacement to check |
|---|---|
| \(|x-a|\), \(x<a\) | \(x-a<0\), so \(|x-a|=-(x-a)\) |
| \(|x-a|\), \(x>a\) | \(x-a>0\), so \(|x-a|=x-a\) |
| \(|g(x)|\), either side | Determine the sign of \(g(x)\) on that side before replacing it |
| Several absolute values | Resolve each enclosed expression independently |
The essential discipline is to attach every replacement to the interval where it is valid. Once that is done, the limit calculation is an ordinary one-sided limit calculation.
Check Your Understanding
For each question, first state the sign of every absolute-value input on the specified side.
- Evaluate \(\displaystyle\lim_{x\to 0^-}\frac{|x|}{x}\).
- Evaluate \(\displaystyle\lim_{x\to 3^+}\frac{|x^2-9|}{x-3}\).
- For \(x\to -2^-\), is \(|x+2|=x+2\) or \(|x+2|=-(x+2)\)? Explain using the sign of \(x+2\).
- Evaluate the left- and right-hand limits of \(\displaystyle\frac{|x-4|}{|x-4|}\) as \(x\to 4\).
- Why is it necessary to check the sign of each enclosed expression separately when a formula contains more than one absolute value?
