Choosing a Method for a Mixed Limit
A mixed limit problem may combine several different issues: the formula can change at a point, a denominator can vanish, a radical can conceal cancellation, or the input can tend to positive or negative infinity. The first task is not to start manipulating symbols. It is to identify which of these features controls the limit.
For a one-sided limit at a finite number, use the formula that applies on the indicated side and determine the signs of factors near the point. For a rational expression, factoring can reveal whether a zero cancels or whether the denominator keeps the quotient unbounded. For a radical difference at infinity, multiplying by a conjugate often exposes the smaller term that remains after the leading terms cancel.
For limits at infinity, \(x\to+\infty\) means that \(x\) increases without bound, while \(x\to-\infty\) means that \(x\) decreases without bound. In either case, the relevant expression must be defined for inputs sufficiently far in the stated direction. A limit of \(+\infty\) or \(-\infty\) describes unbounded behavior; it is not a finite real limit.
A Reliable Decision Process
Before calculating, record the direction of approach and the formula valid in that direction. Then simplify only by identities that hold for nearby inputs in the domain. In particular, cancellation of a common factor is valid away from its zero, even if the original expression is undefined at the limit point. It does not mean the original function is defined there.
For a piecewise formula, select the branch used for inputs on that side. At infinity, distinguish positive from negative inputs.
Check for a vanishing denominator, a common factor, a radical difference, or competing leading terms.
Factor, rationalize, or compare dominant terms as appropriate. Keep track of whether each factor is positive or negative.
Check that the transformed expression is equivalent near the limit, and confirm its sign and limiting value.
Two facts are especially useful in radical problems. First, \(\sqrt{x^2}=|x|\), not always \(x\). Second, for any expressions \(A\) and \(B\), the conjugate identity \((A-B)(A+B)=A^2-B^2\) can eliminate a square root from a difference. The sign hidden in \(|x|\) is essential when \(x\) tends to negative infinity.
Worked Example: A Piecewise Function with Different One-Sided Limits
Worked Example: A Piecewise Boundary and a Radical Quotient
Suppose
Find the left-hand and right-hand limits at \(x=1\), and decide whether the two-sided limit exists.
For \(x\to1^-\), the applicable branch is \((x+1)/(x-1)\). The numerator tends to \(2\), so it is positive for \(x\) sufficiently close to \(1\). The denominator \(x-1\) is negative on the left and tends to zero. Therefore the quotient decreases without bound:
For \(x\to1^+\), use the radical branch. Direct substitution gives a zero numerator and a zero denominator, so first multiply by the conjugate:
For \(x>1\), cancellation of \(x-1\) is valid. Thus, for inputs in this right-hand branch, the quotient equals \(1/(\sqrt{x+3}+2)\). Substituting the limiting input into this simplified expression gives
Check: The conjugate calculation is an exact identity for \(x>1\), and the simplified expression tends to \(1/4\). The left-hand behavior is negative and unbounded, whereas the right-hand behavior approaches a finite number. Since the one-sided limits are not equal, the two-sided limit at \(1\) does not exist. The assigned value \(f(1)=0\) does not change either one-sided limit.
Rational Singularities: Factor First, Then Track Signs
If a rational function has a denominator that tends to zero, substitution alone does not determine the limit. Factoring numerator and denominator can reveal a common factor and leave a simpler expression whose sign is clear. If a factor remains in the denominator, its sign on each side often determines whether the limit is positive or negative infinity.
Worked Example: Opposite Infinite Limits at a Rational Singularity
Evaluate the one-sided limits of
as \(x\to3^-\) and \(x\to3^+\).
Factor the numerator and denominator:
For \(x\ne2,3\), cancellation gives
Near \(3\), the factor \(x+2\) is positive and tends to \(5\). On the left, \(x-3<0\) and approaches zero, so
On the right, \(x-3>0\) and approaches zero, so
Check: At \(x=2.9\), the reduced expression is \(4.9/(-0.1)=-49\), consistent with negative values on the left. At \(x=3.1\), it is \(5.1/0.1=51\), consistent with positive values on the right. These sample values are checks of the signs; the factor analysis establishes the limits. The two-sided limit does not exist because the one-sided behaviors disagree.
Radical Differences at Positive Infinity
When two large terms in a radical difference nearly cancel, their individual leading sizes may not reveal the limit. Multiplication by the conjugate converts the difference into a quotient, often leaving a simpler expression with a finite limit.
Worked Example: A Radical Difference as \(x\to+\infty\)
Evaluate
Rationalize the difference:
Since \(x>0\) as \(x\to+\infty\), divide numerator and denominator by \(x\), using \(\sqrt{4x^2+5x}=x\sqrt{4+5/x}\):
As \(x\to+\infty\), \(5/x\to0\), so the denominator tends to \(\sqrt4+2=4\). Hence the limit is \(5/4\).
Check: The rationalized expression is exactly equal to the original expression wherever the conjugate denominator is nonzero; here that denominator is positive for \(x>0\). Its limit is \(5/4\), confirming that the original difference approaches a positive finite value rather than zero or infinity.
Radicals at Negative Infinity: The Absolute Value Matters
The same technique works at negative infinity, but factoring \(x^2\) from a square root produces \(|x|\). When \(x<0\), \(|x|=-x\). Ignoring this sign can reverse the leading behavior and invalidate the argument.
Worked Example: A Radical Difference as \(x\to-\infty\)
Evaluate
Multiply by the conjugate of the expression. The radicand is positive for all sufficiently negative \(x\), so the expression is defined near this end:
For \(x<0\), \(\sqrt{9x^2+4x}=-x\sqrt{9+4/x}\). Therefore
As \(x\to-\infty\), \(4/x\to0\), and the denominator tends to \(3+3=6\). Thus the limit is \(-4/6=-2/3\).
Check: The exact transformed expression tends to \(-2/3\), and its denominator is positive near negative infinity, so its negative sign is consistent. For example, at \(x=-10\), the original expression is approximately \(\sqrt{860}-30\), which is about \(-0.675\), close to \(-2/3\).
Common Pitfall: Guessing from Leading Terms
A tempting but incorrect argument for the last example is to replace \(\sqrt{9x^2+4x}\) by its leading behavior \(-3x\) as \(x\to-\infty\), then infer that \(-3x+3x=0\) gives the limit. The leading terms do cancel, but that cancellation does not determine the remaining value. The next-order contribution matters, and rationalization shows that the limit is \(-2/3\), not \(0\).
There is also a sign trap in the identity \(\sqrt{x^2}=|x|\). At negative inputs, writing \(\sqrt{x^2}=x\) is false: for example, \(\sqrt{(-5)^2}=5\), not \(-5\). A reliable calculation keeps the absolute value until the direction of approach determines its value.
Check Your Understanding
For each question, identify the relevant side or end before choosing a method.
- If a piecewise function uses different formulas for \(x<a\) and \(x>a\), which formula should be used to compute \(\lim_{x\to a^-}f(x)\)?
- For \(\frac{x+1}{x-4}\), determine the sign of the expression as \(x\to4^-\) and as \(x\to4^+\). What does this imply about the one-sided limits?
- Rationalize \(\sqrt{x^2+6x}-x\) as \(x\to+\infty\), then evaluate the limit.
- Why is \(\sqrt{x^2}=-x\) when \(x\to-\infty\), rather than \(x\)?
- Can a two-sided limit exist if the left-hand limit is \(-\infty\) and the right-hand limit is \(+\infty\)? Explain.
