Why These Limits Need More Than Substitution
When a limit contains a radical difference and a polynomial denominator, direct substitution can produce \(0/0\). That form does not give the limit; it signals that the expression needs to be rewritten. The useful sequence is often to multiply by a conjugate, factor the resulting polynomials, cancel a common factor, and then substitute.
The order matters because rationalizing and factoring reveal factors that may not be visible in the original expression. The goal is not to cancel terms that merely look alike, but to obtain a common factor in a numerator and denominator and simplify the expression for nearby values of \(x\).
A Reliable Sequence of Steps
Suppose the limit is as \(x\) approaches \(a\), and direct substitution makes both numerator and denominator zero. First identify which radical difference is responsible for the zero in the numerator. Multiply by its conjugate, meaning the expression with the same terms and the opposite sign between them. Then simplify using the difference-of-squares identity, factor the resulting polynomial and the denominator, and cancel a shared factor.
Cancellation is valid only for \(x\ne a\) when the cancelled factor is \(x-a\). That is enough for a limit: the simplified expression agrees with the original at all sufficiently close points except possibly \(a\). Once the simplified expression has a denominator that is nonzero at \(a\), direct substitution gives the limit.
Worked Example: Rationalize, Then Factor the Denominator
Worked Example: Rationalize, Then Factor the Denominator
Evaluate \(\displaystyle \lim_{x\to 1}\frac{\sqrt{x+3}-2}{x^2-1}\).
Direct substitution gives \(\sqrt{1+3}-2=0\) in the numerator and \(1^2-1=0\) in the denominator. Multiply the numerator by its conjugate and multiply the denominator by that same conjugate, which is equivalent to multiplying the fraction by \(1\):
Simplify the difference of squares in the numerator and factor the denominator:
For \(x\ne 1\), cancel the common factor \(x-1\). The remaining denominator is nonzero at \(x=1\), so substitute:
Check: The simplified expression tends to \(1/8\), and all transformations were equalities for \(x\ne1\) sufficiently close to \(1\). The original expression is undefined at \(1\), but its limit is \(1/8\).
Worked Example: Rationalize Two Radicals and Factor
Worked Example: Rationalize Two Radicals and Factor
Evaluate \(\displaystyle \lim_{x\to 2}\frac{\sqrt{x+3}-\sqrt{2x+1}}{x^2-4}\).
At \(x=2\), both radicands equal \(5\), so the numerator is zero. The denominator is also \(2^2-4=0\). The conjugate of the numerator is \(\sqrt{x+3}+\sqrt{2x+1}\). Multiplying by the conjugate over itself gives:
The new numerator is \(x+3-2x-1=2-x=-(x-2)\). Factor the denominator using the difference of squares:
For \(x\ne2\), cancel \(x-2\), keeping the negative sign. Now substitute \(x=2\):
Check: The denominator in the simplified expression is \(4(2\sqrt5)=8\sqrt5\) at \(x=2\), so the value, including the negative sign, is \(-1/(8\sqrt5)\).
Worked Example: Apply Conjugates in Stages
Worked Example: Apply Conjugates in Stages
Evaluate \(\displaystyle \lim_{x\to 2}\frac{\sqrt{2+\sqrt{x+2}}-2}{x-2}\).
At \(x=2\), the inner radical is \(\sqrt4=2\), the outer radical is \(\sqrt{2+2}=2\), and the denominator is zero. First use the conjugate of the outer radical difference:
The numerator still contains a difference of square roots. Multiply by its conjugate, \(\sqrt{x+2}+2\), over itself:
The new numerator is \(x-2\), which cancels with the denominator factor for \(x\ne2\). Substitute into the remaining expression:
Check: At \(x=2\), each parenthesized factor in the simplified denominator equals \(4\), so their product is \(16\). The limit is \(1/16\).
Why the Steps Preserve the Limit
The algebra in these examples works for two reasons. First, the identity \((u-v)(u+v)=u^2-v^2\) removes a radical difference when \(u\) and \(v\) are square roots. Second, after simplifying that difference and factoring, cancellation makes the expression easier to evaluate.
For real \(u\) and \(v\), \((u-v)(u+v)=u^2-v^2\). Thus multiplying by the conjugate replaces a difference of square roots by the difference of their radicands.
Rewrite any resulting polynomial and the original denominator as products. A shared factor can then be identified without cancelling across addition or subtraction.
For \(x\ne a\) sufficiently close to \(a\), the multipliers are defined and nonzero and any cancelled factor is nonzero. The simplified expression therefore equals the original there.
If the simplified expression is continuous at \(a\) and its denominator is nonzero there, its limit is its value at \(a\). Since it equals the original expression for nearby \(x\ne a\), that value is also the original limit.
More explicitly, if the simplified expression is \(g(x)\), then \(f(x)=g(x)\) for all \(x\ne a\) in some interval around \(a\). For any required tolerance in the definition of the limit, continuity of \(g\) supplies a sufficiently small distance from \(a\) that makes \(g(x)\) close to \(g(a)\). The equality \(f(x)=g(x)\) on that punctured interval then makes \(f(x)\) just as close to \(g(a)\). This is why simplifying away a zero factor does not change the limit.
A Common Cancellation Mistake
In the first worked example, after rationalizing and factoring, the expression is
A wrong calculation cancels \(x-1\) and then also discards the remaining factor \(x+1\), giving \(1/(\sqrt{x+3}+2)\) and the incorrect limit \(1/4\). The correction is to cancel only the common factor \(x-1\):
The factor \(x+1\) remains in the denominator, so the limit is \(1/(2\cdot4)=1/8\). Factoring completely before cancelling makes the remaining factors visible.
When This Combination Is Useful
This approach is especially useful when substitution gives \(0/0\), the numerator contains a difference of radicals, and the denominator is a polynomial with a factor that may match the polynomial created by rationalization. In nested expressions, it may take more than one conjugate multiplication to expose that factor.
Keep the algebra tied to its conditions: check that the radicals are defined for real \(x\) near the point of approach; ensure each conjugate used as a divisor is nonzero nearby; factor before cancelling; and verify that the final denominator is nonzero at the point. If those checks hold, substitution into the simplified expression completes the limit.
Check Your Understanding
For each question, show the conjugate multiplication and factorization where needed, and state why cancellation is valid near the limit point.
- Evaluate \(\displaystyle \lim_{x\to 3}\frac{\sqrt{x+6}-3}{x^2-9}\).
- For \(\displaystyle \frac{\sqrt{x+4}-\sqrt{3x}}{x-2}\), identify the conjugate and the polynomial obtained in the numerator after multiplication.
- Why does cancelling a factor \(x-a\) require considering \(x\ne a\), and why can that still be used to evaluate a limit at \(a\)?
- In \(\displaystyle \frac{x-1}{(x-1)(x+1)(\sqrt{x+3}+2)}\), which factors remain after valid cancellation?
- For a nested radical difference, why might one conjugate multiplication leave another radical difference to simplify?
