Limits Using Conjugates
In our study of limits involving square roots and other radicals, we have already seen that direct substitution often leads to the indeterminate form ##\frac{0}{0}##. One of the most effective ways to handle such expressions is to use conjugates. In this reading, we focus specifically on limits using conjugates in calculus, organise the basic strategy, and then practise it through a wide range of fully solved examples.
1. Conjugates and the difference of squares identity
Whenever we see expressions like ##\sqrt{A} – \sqrt{B}## or ##\sqrt{A} + \sqrt{B}##, we can form the conjugate by changing the sign in the middle:
- The conjugate of ##\sqrt{A} – \sqrt{B}## is ##\sqrt{A} + \sqrt{B}##.
- The conjugate of ##\sqrt{A} + \sqrt{B}## is ##\sqrt{A} – \sqrt{B}##.
The key identity is the difference of squares:
###(\sqrt{A} – \sqrt{B})(\sqrt{A} + \sqrt{B}) = A – B.###
This identity lies at the heart of limits using conjugates in calculus. By multiplying a surd-based expression by a suitable conjugate, we remove square roots and obtain a simpler algebraic expression where standard techniques apply.
2. General strategy for limits using conjugates
When we encounter a limit with square roots and direct substitution gives ##\frac{0}{0}##, we usually follow these steps:
- Apply direct substitution to identify the indeterminate form.
- Locate the surd expression that is causing the difficulty (often a difference of square roots).
- Multiply numerator and denominator by the conjugate of that surd expression.
- Use the identity to remove square roots and simplify.
- Cancel common factors where possible.
- Evaluate the limit of the simplified expression, often by direct substitution.
We now apply this routine in detail to many examples so that the technique becomes automatic.
3. Classic examples: roots in the numerator
Example 1
Evaluate ###\lim_{x \to 4} \frac{\sqrt{x} – 2}{x – 4}.###
Solution. Direct substitution gives
###\frac{\sqrt{4} – 2}{4 – 4} = \frac{2 – 2}{0} = \frac{0}{0}.###
We multiply numerator and denominator by the conjugate ##\sqrt{x} + 2##:
###\frac{\sqrt{x} – 2}{x – 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}.###
The numerator becomes
###(\sqrt{x} – 2)(\sqrt{x} + 2) = x – 4.###
So for ##x \neq 4##,
###\frac{x – 4}{(x – 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}.###
Now the limit is
###\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{2 + 2} = \frac{1}{4}.###
Example 2
Evaluate ###\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x}.###
Solution. Direct substitution gives ##\frac{0}{0}##. We use the conjugate ##\sqrt{1 + x} + 1##:
###\frac{\sqrt{1 + x} – 1}{x} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}.###
The numerator simplifies to
###(1 + x) – 1 = x.###
So for ##x \neq 0##,
###\frac{x}{x(\sqrt{1 + x} + 1)} = \frac{1}{\sqrt{1 + x} + 1}.###
Taking the limit,
###\lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{1 + 1} = \frac{1}{2}.###
Example 3
Evaluate ###\lim_{x \to 9} \frac{\sqrt{x} – 3}{x – 9}.###
Solution. Substitution gives ##\frac{0}{0}##. Multiply by the conjugate ##\sqrt{x} + 3##:
###\frac{\sqrt{x} – 3}{x – 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3}.###
The numerator becomes ##x – 9##, so
###\frac{x – 9}{(x – 9)(\sqrt{x} + 3)} = \frac{1}{\sqrt{x} + 3}, \quad x \neq 9.###
Thus
###\lim_{x \to 9} \frac{\sqrt{x} – 3}{x – 9} = \frac{1}{3 + 3} = \frac{1}{6}.###
Example 4
Evaluate ###\lim_{x \to a} \frac{\sqrt{x} – \sqrt{a}}{x – a}, \quad a > 0.###
Solution. This is a general form of the previous example. Direct substitution gives ##\frac{0}{0}##. Multiply by the conjugate ##\sqrt{x} + \sqrt{a}##:
###\frac{\sqrt{x} – \sqrt{a}}{x – a} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}.###
The numerator becomes ##x – a##. Hence
###\frac{x – a}{(x – a)(\sqrt{x} + \sqrt{a})} = \frac{1}{\sqrt{x} + \sqrt{a}}, \quad x \neq a.###
Now
###\lim_{x \to a} \frac{\sqrt{x} – \sqrt{a}}{x – a} = \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}.###
Example 5
Evaluate ###\lim_{x \to 0} \frac{\sqrt{1 + 2x} – 1}{x}.###
Solution. Multiply numerator and denominator by the conjugate ##\sqrt{1 + 2x} + 1##:
###\frac{\sqrt{1 + 2x} – 1}{x} \cdot \frac{\sqrt{1 + 2x} + 1}{\sqrt{1 + 2x} + 1}.###
The numerator becomes
###(1 + 2x) – 1 = 2x.###
So
###\frac{2x}{x(\sqrt{1 + 2x} + 1)} = \frac{2}{\sqrt{1 + 2x} + 1}, \quad x \neq 0.###
Now the limit is
###\lim_{x \to 0} \frac{2}{\sqrt{1 + 2x} + 1} = \frac{2}{1 + 1} = 1.###
4. Roots appearing in both terms
Example 6
Evaluate ###\lim_{x \to 0} \frac{\sqrt{4 + x} – \sqrt{4 – x}}{x}.###
Solution. Substitution gives ##\frac{0}{0}##. We use the conjugate of the numerator, ##\sqrt{4 + x} + \sqrt{4 – x}##:
###\frac{\sqrt{4 + x} – \sqrt{4 – x}}{x} \cdot \frac{\sqrt{4 + x} + \sqrt{4 – x}}{\sqrt{4 + x} + \sqrt{4 – x}}.###
The numerator becomes
###(4 + x) – (4 – x) = 2x.###
So
###\frac{2x}{x(\sqrt{4 + x} + \sqrt{4 – x})} = \frac{2}{\sqrt{4 + x} + \sqrt{4 – x}}, \quad x \neq 0.###
Now
###\lim_{x \to 0} \frac{2}{\sqrt{4 + x} + \sqrt{4 – x}} = \frac{2}{2 + 2} = \frac{1}{2}.###
Example 7
Evaluate ###\lim_{x \to 0} \frac{\sqrt{a + x} – \sqrt{a – x}}{x}, \quad a > 0.###
Solution. Multiply by the conjugate ##\sqrt{a + x} + \sqrt{a – x}##:
###\frac{\sqrt{a + x} – \sqrt{a – x}}{x} \cdot \frac{\sqrt{a + x} + \sqrt{a – x}}{\sqrt{a + x} + \sqrt{a – x}}.###
The numerator becomes
###(a + x) – (a – x) = 2x.###
So, for ##x \neq 0##,
###\frac{2x}{x(\sqrt{a + x} + \sqrt{a – x})} = \frac{2}{\sqrt{a + x} + \sqrt{a – x}}.###
Taking the limit,
###\lim_{x \to 0} \frac{2}{\sqrt{a + x} + \sqrt{a – x}} = \frac{2}{\sqrt{a} + \sqrt{a}} = \frac{1}{\sqrt{a}}.###
5. Rationalising the denominator
Sometimes the square root is in the denominator, and we want to remove it to simplify a limit. Conjugates are again the main tool.
Example 8
Evaluate ###\lim_{x \to 0} \frac{x}{\sqrt{1 + x} – 1}.###
Solution. Direct substitution gives ##\frac{0}{0}##. Multiply numerator and denominator by the conjugate ##\sqrt{1 + x} + 1##:
###\frac{x}{\sqrt{1 + x} – 1} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}.###
The denominator becomes
###(1 + x) – 1 = x.###
Hence, for ##x \neq 0##,
###\frac{x(\sqrt{1 + x} + 1)}{x} = \sqrt{1 + x} + 1.###
The limit is
###\lim_{x \to 0} (\sqrt{1 + x} + 1) = 1 + 1 = 2.###
Example 9
Evaluate ###\lim_{x \to 0} \frac{x}{\sqrt{1 – x} – 1}.###
Solution. Again we have ##\frac{0}{0}##. Multiply by ##\sqrt{1 – x} + 1##:
###\frac{x}{\sqrt{1 – x} – 1} \cdot \frac{\sqrt{1 – x} + 1}{\sqrt{1 – x} + 1}.###
The denominator becomes
###(1 – x) – 1 = -x.###
Thus, for ##x \neq 0##,
###\frac{x(\sqrt{1 – x} + 1)}{-x} = -(\sqrt{1 – x} + 1).###
Now
###\lim_{x \to 0} -(\sqrt{1 – x} + 1) = -(1 + 1) = -2.###
Example 10
Evaluate ###\lim_{x \to 0} \frac{x}{\sqrt{4 + x} – 2}.###
Solution. Multiply numerator and denominator by ##\sqrt{4 + x} + 2##:
###\frac{x}{\sqrt{4 + x} – 2} \cdot \frac{\sqrt{4 + x} + 2}{\sqrt{4 + x} + 2}.###
The denominator becomes
###(4 + x) – 4 = x.###
So for ##x \neq 0##,
###\frac{x(\sqrt{4 + x} + 2)}{x} = \sqrt{4 + x} + 2.###
Hence
###\lim_{x \to 0} (\sqrt{4 + x} + 2) = 2 + 2 = 4.###
6. More involved examples
Example 11
Evaluate ###\lim_{x \to 0} \frac{\sqrt{x^2 + 4} – 2}{x^2}.###
Solution. Direct substitution gives ##\frac{0}{0}##. Multiply numerator and denominator by the conjugate ##\sqrt{x^2 + 4} + 2##:
###\frac{\sqrt{x^2 + 4} – 2}{x^2} \cdot \frac{\sqrt{x^2 + 4} + 2}{\sqrt{x^2 + 4} + 2}.###
The numerator becomes
###(x^2 + 4) – 4 = x^2.###
Thus, for ##x \neq 0##,
###\frac{x^2}{x^2(\sqrt{x^2 + 4} + 2)} = \frac{1}{\sqrt{x^2 + 4} + 2}.###
Now
###\lim_{x \to 0} \frac{1}{\sqrt{x^2 + 4} + 2} = \frac{1}{\sqrt{0 + 4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}.###
Example 12
Evaluate ###\lim_{x \to 0} \frac{\sqrt{2x + 3} – \sqrt{3}}{x}.###
Solution. Multiplying by the conjugate ##\sqrt{2x + 3} + \sqrt{3}## gives
###\frac{\sqrt{2x + 3} – \sqrt{3}}{x} \cdot \frac{\sqrt{2x + 3} + \sqrt{3}}{\sqrt{2x + 3} + \sqrt{3}}.###
The numerator becomes
###(2x + 3) – 3 = 2x.###
So, for ##x \neq 0##,
###\frac{2x}{x(\sqrt{2x + 3} + \sqrt{3})} = \frac{2}{\sqrt{2x + 3} + \sqrt{3}}.###
Now
###\lim_{x \to 0} \frac{2}{\sqrt{2x + 3} + \sqrt{3}} = \frac{2}{\sqrt{3} + \sqrt{3}} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}.###
Example 13
Evaluate ###\lim_{x \to a} \frac{\sqrt{1 + x} – \sqrt{1 + a}}{x – a}.###
Solution. This generalises Example 2. Multiply by the conjugate ##\sqrt{1 + x} + \sqrt{1 + a}##:
###\frac{\sqrt{1 + x} – \sqrt{1 + a}}{x – a} \cdot \frac{\sqrt{1 + x} + \sqrt{1 + a}}{\sqrt{1 + x} + \sqrt{1 + a}}.###
The numerator becomes
###(1 + x) – (1 + a) = x – a.###
So
###\frac{x – a}{(x – a)(\sqrt{1 + x} + \sqrt{1 + a})} = \frac{1}{\sqrt{1 + x} + \sqrt{1 + a}}, \quad x \neq a.###
Taking the limit,
###\lim_{x \to a} \frac{\sqrt{1 + x} – \sqrt{1 + a}}{x – a} = \frac{1}{\sqrt{1 + a} + \sqrt{1 + a}} = \frac{1}{2\sqrt{1 + a}}.###
Example 14
Evaluate ###\lim_{x \to 0} \left( \frac{\sqrt{1 + x} – 1}{x} – \frac{1}{2} \right).###
Solution. From Example 2 we know
###\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x} = \frac{1}{2}.###
Therefore, the limit of the difference is
###\lim_{x \to 0} \left( \frac{\sqrt{1 + x} – 1}{x} – \frac{1}{2} \right) = \frac{1}{2} – \frac{1}{2} = 0.###
This shows how conjugate-based simplifications can be used inside more complicated expressions.
Example 15
Evaluate ###\lim_{h \to 0} \frac{\sqrt{x + h} – \sqrt{x}}{h}, \quad x > 0.###
Solution. This is a difference quotient for the square-root function. Multiply numerator and denominator by the conjugate ##\sqrt{x + h} + \sqrt{x}##:
###\frac{\sqrt{x + h} – \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}.###
The numerator becomes
###(x + h) – x = h.###
Thus, for ##h \neq 0##,
###\frac{h}{h(\sqrt{x + h} + \sqrt{x})} = \frac{1}{\sqrt{x + h} + \sqrt{x}}.###
Now
###\lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}.###
This result is exactly the derivative of ##\sqrt{x}##, obtained using limits using conjugates in calculus.
7. Summary
In this reading, we have treated limits using conjugates in calculus as a focused technique:
- Conjugates such as ##\sqrt{A} – \sqrt{B}## and ##\sqrt{A} + \sqrt{B}## allow us to remove square roots using the identity ###(\sqrt{A} – \sqrt{B})(\sqrt{A} + \sqrt{B}) = A – B###.
- When direct substitution in a limit produces ##\frac{0}{0}## and square roots are present, multiplying by a conjugate often reveals a hidden factor that can be cancelled.
- We saw how to handle surds in the numerator, surds in the denominator, and symmetric expressions with two square roots.
- General formulas such as ###\lim_{x \to a} \frac{\sqrt{x} – \sqrt{a}}{x – a} = \frac{1}{2\sqrt{a}}### and ###\lim_{h \to 0} \frac{\sqrt{x + h} – \sqrt{x}}{h} = \frac{1}{2\sqrt{x}}### arise naturally from this method and later reappear in differentiation.
With a firm command over limits using conjugates in calculus, we are ready to tackle more advanced problems involving radicals and to connect these ideas with the study of derivatives and continuous functions.
8. References for further exploration
- The general discussion of limits and algebraic techniques at Wikipedia includes many examples with square roots.
- Lecture notes and problem sets from institutions and organisations such as the American Mathematical Society (AMS) and the National Institute of Standards and Technology (NIST) often feature rationalisation and conjugate methods in the context of limits.
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