Why Signs Matter in Infinite Limits
When a function becomes unbounded near an input, its sign determines which infinite limit it may have. A function growing without bound through positive values tends to \(+\infty\); one decreasing without bound through negative values tends to \(-\infty\). A sign error can therefore change the answer, hide different behavior from the two sides, or lead to a limit that does not exist.
The direction of approach matters. So do the parity of a power, the sign of a rational expression’s denominator, and the definition of the square root. A reliable calculation keeps these pieces separate: first identify the domain and direction, then determine signs, and only then state the limit.
What an Infinite Limit Means
Let \(f:D\to\mathbb R\), where \(D\) is a subset of the real numbers, and let \(a\) be finite. For a two-sided limit, \(a\) must be an accumulation point of \(D\): every interval around \(a\) must contain a point of \(D\) other than \(a\). This prevents a limit statement from being true merely because there are no domain inputs near \(a\).
For a one-sided limit, the domain must have inputs arbitrarily close to \(a\) from the specified side. For example, \(a\) is a left accumulation point of \(D\) if every interval \((a-\delta,a)\), with \(\delta>0\), contains a point of \(D\). The left-hand definition uses only inputs \(x\in D\) with \(a-\delta<x<a\); the right-hand definition uses only inputs \(x\in D\) with \(a<x<a+\delta\). The same inequalities \(f(x)>M\) and \(f(x)<-M\) define the respective limits to \(+\infty\) and \(-\infty\).
At infinity, \(\lim_{x\to+\infty}f(x)=+\infty\) means that for every \(M>0\), there is an \(R\) such that \(x\in D\) and \(x>R\) imply \(f(x)>M\). The domain must be unbounded above. For \(x\to-\infty\), the domain must be unbounded below, and the condition is applied to inputs \(x\in D\) with \(x<R\). Replacing \(f(x)>M\) by \(f(x)<-M\) defines a limit of \(-\infty\).
The Sign Rules Behind One-Sided Limits
Write \(h=x-a\) when studying a finite input \(a\). On the left, \(h<0\); on the right, \(h>0\). For an even positive integer \(n\), \(h^n>0\) on both sides. For an odd positive integer \(n\), \(h^n\) has the same sign as \(h\). These facts explain why a denominator with an even power can produce the same infinite limit from both sides, while an odd power can produce opposite signs.
Here is a direct justification of the unboundedness assertion. Given \(M>0\), choose a neighborhood where the stated bound on \(u\) holds, and restrict further to \(|x-a|<(c/M)^{1/n}\). Then \(|x-a|^n<c/M\), so
The fixed sign of the quotient then tells us whether the values exceed \(M\) or are less than \(-M\). The domain must also contain inputs arbitrarily close on each side for the corresponding one-sided limit to be considered.
Worked Example: An Odd Power Gives Opposite One-Sided Signs
Worked Example: An Odd Power Gives Opposite One-Sided Signs
Find the one-sided infinite limits of \(f(x)=\frac{x+1}{(x-2)^3}\) as \(x\to2\). The domain excludes \(x=2\), but it contains inputs arbitrarily close to 2 on both sides.
Take \(|x-2|<1\). Then \(1<x<3\), so \(2<x+1<4\): the numerator is positive and greater than 2. On the left of 2, \(x-2<0\), so \((x-2)^3<0\) and the quotient is negative. On the right, \(x-2>0\), so the quotient is positive.
To verify that the magnitudes grow without bound, let \(M>0\) and choose \(\delta=\min(1,(2/M)^{1/3})\). For \(2-\delta<x<2\), the numerator is greater than 2 and \(0<|x-2|<\delta\). Therefore
For \(2<x<2+\delta\), the numerator is again greater than 2, and
Thus \(\lim_{x\to2^-}f(x)=-\infty\), while \(\lim_{x\to2^+}f(x)=+\infty\). The signs check against the expression: a positive numerator divided by a negative cube is negative on the left, and divided by a positive cube is positive on the right.
Worked Example: A Negative Input Does Not Make a Square Root Negative
Worked Example: A Negative Input Does Not Make a Square Root Negative
Find \(\lim_{x\to-\infty}\bigl(\sqrt{x^2+9}-x\bigr)\). The square root denotes the principal square root, which is always nonnegative.
For every real \(x\), \(x^2+9>x^2\), so \(\sqrt{x^2+9}>|x|\). When \(x<0\), \(|x|=-x\). Consequently,
Given \(M>0\), choose \(R=-M/2\). If \(x<R\), then \(x<-M/2\), so \(-2x>M\). It follows that \(\sqrt{x^2+9}-x>M\). The domain is all real numbers and is unbounded below, so the limit is defined and
As a numerical sign check, at \(x=-5\) the expression is \(\sqrt{34}+5\), which is positive and greater than 10 because \(\sqrt{34}>5\). Replacing \(\sqrt{x^2}\) by \(x\) for negative \(x\) would be invalid: in fact, \(\sqrt{x^2}=|x|\), which equals \(-x\) when \(x<0\).
Worked Example: Check the Numerator and Denominator Separately
Worked Example: Check the Numerator and Denominator Separately
Find the one-sided infinite limits of \(g(x)=\frac{x^2+1}{x+2}\) as \(x\to-2\). The function is undefined at \(-2\), but its domain contains inputs arbitrarily close to \(-2\) on both sides.
If \(|x+2|<1\), then \(-3<x<-1\), so \(x^2+1>2\). The numerator is positive on both sides. For \(x<-2\), the denominator \(x+2\) is negative, so \(g(x)<0\). For \(x>-2\), the denominator is positive, so \(g(x)>0\).
To verify the magnitude, let \(M>0\) and set \(\delta=\min(1,2/M)\). For \(-2-\delta<x<-2\), the numerator is greater than 2 and \(0<|x+2|<\delta\), giving
For \(-2<x<-2+\delta\), the numerator is greater than 2 and \(0<x+2<\delta\), so
Therefore \(\lim_{x\to-2^-}g(x)=-\infty\) and \(\lim_{x\to-2^+}g(x)=+\infty\). The numerator’s positivity and the denominator’s changing sign confirm both conclusions.
Even Powers and the Two-Sided Limit
An even power does not change sign across its zero. This often makes the two-sided infinite limit agree, provided the numerator’s sign is fixed and nonzero near the input. For example, consider \(p(x)=\frac{1}{(x-4)^2}\). For every \(x\ne4\), the denominator is positive, so \(p(x)>0\). Given \(M>0\), if \(0<|x-4|<1/\sqrt M\), then \((x-4)^2<1/M\), and hence \(p(x)>M\). The domain has inputs arbitrarily close to 4, so
The check is immediate: on either side of 4, the squared denominator is positive, and making \(x\) close to 4 makes that positive denominator small. Its reciprocal is therefore positive and unbounded. An even power alone does not guarantee \(+\infty\); a negative numerator would reverse the sign.
A Common Mistake: Using the Right-Side Sign on Both Sides
A frequent error is to notice that \((x-a)^3\) is positive when \(x>a\), then use that sign for a two-sided limit. For the function \(\frac{1}{(x-1)^3}\), the incorrect calculation might be:
The first implication is only valid when \(x\to1^+\). When \(x\to1^-\), \(x-1<0\), so \((x-1)^3<0\), and the reciprocal is negative. In fact, the left-hand limit is \(-\infty\), while the right-hand limit is \(+\infty\). Since the two one-sided limits disagree, the two-sided limit is neither \(+\infty\) nor \(-\infty\).
The correction is to write the direction before deciding the sign. Do not infer a two-sided limit merely from what happens on one side.
A Reliable Sign Audit
Before concluding that an infinite limit is positive or negative, use the following checks. They are particularly useful for rational expressions and radical expressions, where one overlooked sign can change the conclusion.
Identify excluded inputs and confirm that domain points approach from the side or sides under consideration.
For a finite input \(a\), record whether \(x<a\), \(x>a\), or both. For a limit at infinity, record whether inputs go to \(+\infty\) or \(-\infty\).
For a power, use even or odd parity. For a quotient, inspect numerator and denominator separately.
Use an inequality, or a bound away from zero for the numerator, to justify that the function exceeds any prescribed positive threshold in magnitude.
If the two one-sided signs differ, do not assign a single two-sided infinite limit.
Why These Checks Are Useful
The symbol \(+\infty\) describes values that are positive and exceed every positive bound; \(-\infty\) describes values that are negative and whose absolute values exceed every positive bound. Neither symbol is an ordinary real number. That is why a sign check and a magnitude argument both matter: sign alone does not establish unbounded growth, and growth in magnitude alone does not establish whether the limit is positive or negative.
In practice, factoring a denominator can expose the factors that change sign, and a quick test input on each side can help catch an error. Such tests are checks, not proofs: the conclusion still requires reasoning that applies throughout a sufficiently small neighborhood. The inequalities in the examples supply that reasoning.
Check Your Understanding
For each question, identify the direction and sign before deciding what limit statement is justified.
- For \(f(x)=\frac{1}{(x+3)^4}\), what are the left-hand and right-hand limits as \(x\to-3\), and why is the denominator positive on both sides?
- For \(g(x)=\frac{1}{(x+3)^5}\), how do the left-hand and right-hand signs differ as \(x\to-3\)?
- For \(h(x)=\frac{x^2+4}{x-1}\), what sign does the function have just to the left of \(1\), and what sign does it have just to the right?
- When \(x<0\), what is the correct relationship between \(\sqrt{x^2}\) and \(x\)?
- Why must a definition of a one-sided infinite limit specify that domain inputs occur arbitrarily close from the chosen side?
