Why Limit Calculations Need Care
A limit describes how the values of a function behave near an input, not necessarily what the function equals at that input. This distinction makes limits flexible: changing or omitting a function’s value at one point does not change its limit there.
But the algebra used to calculate limits still has conditions. In particular, the quotient law requires that the denominator approach a nonzero number. Cancellation is valid for common factors, not for terms that merely appear in a sum. And direct substitution is a useful test, not a universal method.
This tutorial develops a careful way to identify these issues. The aim is not to avoid algebraic shortcuts, but to know exactly when they preserve the limit.
The Quotient Law Has a Condition
The quotient law does not say that every quotient limit can be found by dividing the limits. It says this is valid when the denominator approaches a nonzero number. When the denominator approaches zero, direct substitution may produce a form such as \(0/0\). That is not a limit value; it means this law cannot determine the answer. More analysis is needed, and the limit might exist or might not.
Here is why the nonzero condition matters. If \(g(x)\to B\ne 0\), then for inputs sufficiently close to \(a\), \(g(x)\) stays away from zero. In fact, it can be made close enough to \(B\) that \(|g(x)|>|B|/2\). Thus the reciprocals \(1/g(x)\) are defined nearby and can be compared with \(1/B\).
Why the Quotient Law Works
The following argument proves that \(g(x)\to B\ne 0\) implies \(1/g(x)\to 1/B\). Once this reciprocal limit is established, the product law gives the quotient law.
Let \(\varepsilon>0\) be given. Because \(g(x)\to B\), we can require \(|g(x)-B|<\min(|B|/2,\varepsilon B^2/2)\) for all \(x\) sufficiently close to \(a\), with \(x\ne a\).
The triangle inequality gives \(|g(x)|\ge |B|-|g(x)-B|>|B|/2\). In particular, \(g(x)\ne 0\) for these inputs.
For these inputs, \(\left|\frac{1}{g(x)}-\frac{1}{B}\right|=\frac{|g(x)-B|}{|g(x)||B|}<\frac{2|g(x)-B|}{B^2}<\varepsilon\). Therefore \(1/g(x)\to 1/B\).
Now if \(f(x)\to A\) as well, then \(f(x)/g(x)=f(x)(1/g(x))\) wherever the quotient is defined nearby. The product law gives the limit \(A(1/B)=A/B\). This reasoning fails when \(B=0\): the lower bound keeping \(g(x)\) away from zero is no longer available.
Worked Example: A Zero Denominator Limit Is Not an Answer
Worked Example: A Zero Denominator Limit Is Not an Answer
Find \(\lim_{x\to 3}\frac{x^2-9}{x-3}\).
Direct substitution gives a numerator approaching \(3^2-9=0\) and a denominator approaching \(3-3=0\). The resulting form \(0/0\) does not give the limit, and the quotient law is not applicable because the denominator limit is zero.
For \(x\ne 3\), factor the numerator and cancel the common factor:
The original quotient and \(x+3\) agree for every \(x\ne 3\), which includes all nearby inputs in the limit except the point \(x=3\). Therefore,
Check: for every \(x\ne 3\), the simplified expression \(x+3\) equals the original quotient, so both have the same limiting value, \(6\). The original quotient is undefined at \(x=3\); the limit is still \(6\).
This example illustrates a useful principle: expressions that agree at every sufficiently nearby input except possibly the point of approach have the same limit there. Cancellation is legitimate here because \(x-3\) is a common factor, and the cancellation is made only for \(x\ne 3\).
Cancellation Is for Factors, Not Terms
A factor multiplies an entire expression. A term is one part of a sum or difference. A common factor can be canceled from numerator and denominator when it is nonzero at the inputs under consideration. Separate terms cannot be canceled across a fraction bar. For example, a sum \(u+v\) is not generally divisible by \(u\) in a way that lets \(u\) be canceled against another \(u\) below the bar.
Worked Example: Do Not Cancel Across Sums
Worked Example: Do Not Cancel Across Sums
Evaluate \(\lim_{x\to 1}\frac{x^2+1}{x+1}\), and compare it with an invalid cancellation.
At \(x=1\), the denominator approaches \(1+1=2\), which is nonzero. The numerator approaches \(1^2+1=2\). The quotient law therefore applies:
A mistaken calculation might cancel the \(x\) in \(x^2\) against the \(x\) in \(x+1\), then claim that the expression becomes \(x+1\). That step is invalid: \(x+1\) is a sum, not a product with \(x\) as a factor. Substituting \(x=1\) into the incorrect result gives \(1+1=2\), contradicting the correct value \(1\).
Check: the original quotient is continuous at \(x=1\), since its denominator is nonzero there. Its value at \(1\) is \(2/2=1\), in agreement with the limit. The proposed cancellation would not preserve the expression: at \(x=1\), it would change the value from \(1\) to \(2\).
A safe test is to ask whether the numerator and denominator have a shared multiplicative factor. If the alleged cancellation crosses a plus or minus sign, rewrite and verify the algebra before proceeding.
Substitution Is a Test, Not a Universal Rule
Direct substitution works when the function’s limit can be evaluated from the values of its component functions and the relevant limit laws apply. For a quotient, in particular, the denominator must approach a nonzero number. When substitution instead produces \(0/0\), it has identified a blockage—not supplied an answer.
Earlier material on sine limits establishes \(\lim_{u\to 0}\frac{\sin u}{u}=1\). A change of variable can use that result when direct substitution into a trigonometric quotient gives \(0/0\).
Worked Example: A \(0/0\) Substitution Needs More Work
Worked Example: A \(0/0\) Substitution Needs More Work
Find \(\lim_{x\to 0}\frac{\sin(5x)}{x}\).
Substituting \(x=0\) into numerator and denominator gives \(\sin(0)=0\) and \(0\), so direct substitution produces \(0/0\). Since the denominator limit is zero, the quotient law cannot be used to divide the two substituted values.
For \(x\ne 0\), multiply and divide by \(5\) to put the expression in the form of the known sine quotient:
As \(x\to 0\), the input \(5x\to 0\). Using the sine quotient limit gives
Check: writing \(u=5x\) gives \(\sin(5x)/x=5\sin(u)/u\) for \(x\ne0\), and \(u\to0\). The established sine quotient limit makes the latter expression tend to \(5\), as found.
The substitution \(x=0\) was useful because it showed which method would not work. It did not settle the limit. After rewriting, the result came from a limit law that applies to the new expression and a known limit.
A Function Value Is Not Its Limit
The notation \(\lim_{x\to a}f(x)\) concerns values for inputs close to \(a\) with \(x\ne a\). The function value \(f(a)\), when it exists, concerns the single input \(a\). The two can be different, and the limit can exist even if \(f(a)\) is undefined.
Worked Example: The Limit Can Differ from the Function Value
Worked Example: The Limit Can Differ from the Function Value
Suppose
For \(x\ne 1\), factor and cancel the common factor \(x-1\):
Thus, near \(1\) but not at \(1\), \(f(x)=x+1\). Therefore,
But the piecewise definition gives \(f(1)=10\). Hence \(\lim_{x\to1}f(x)=2\), whereas \(f(1)=10\). Check: the limit calculation uses the values \(x+1\) for \(x\ne1\), which approach \(2\); substituting \(x=1\) into the separate point-value clause gives \(10\). These are different questions, so the different answers are consistent.
A Reliable Check Before Using a Limit Law
Before simplifying or substituting, identify what the proposed step assumes. In particular, for a quotient, compute the limit of the denominator separately. If it is nonzero, the quotient law allows division of the limits. If it is zero, pause: the quotient law gives no conclusion. Then check whether an algebraic rewrite is valid for nearby inputs and whether any cancellation is of a genuine common factor.
Check Your Understanding
For each question, identify which limit rule or algebraic principle is relevant before calculating.
- For \(\lim_{x\to 2}\frac{x^2-4}{x-2}\), what does direct substitution produce, and why is that not the answer?
- Why is canceling \(x\) in \(\frac{x+2}{x+3}\) invalid?
- What condition on \(\lim_{x\to a}g(x)\) is needed to use the quotient law for \(f(x)/g(x)\)?
- Can a function’s limit at a point differ from its value at that point? Explain briefly.
- If direct substitution gives \(0/0\), what has been established about the limit?
