In previous sections on limits, we focused mainly on direct substitution and factoring. Those methods work very well for polynomials and many rational functions. However, when square roots or other radicals (surds) appear in a limit, direct substitution often produces the indeterminate form ##\frac{0}{0}## and simple factoring is no longer enough. In such situations, a key algebraic technique is to use limits using rationalisation in calculus.
In this reading, we will see how multiplying by a suitable conjugate helps us remove square roots from the numerator or denominator, turn surds into simpler algebraic expressions, and then evaluate the limit. We will work through many fully solved examples so that the pattern becomes automatic.
1. Why Surds Cause Trouble in Limits
Consider the limit
###\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x}.###
If we try direct substitution, we get
###\frac{\sqrt{1 + 0} – 1}{0} = \frac{1 – 1}{0} = \frac{0}{0},###
which is an indeterminate form. We cannot decide the limit from this alone. Unlike a pure polynomial, the expression contains a square root, so standard factoring does not apply directly. This is typical for limits involving surds: the root hides a factor that we need to expose.
The idea of limits using rationalisation in calculus is to multiply the expression by a carefully chosen conjugate so that the square roots disappear from the numerator or denominator. After that, we usually find that the problematic factor can be cancelled, and the limit becomes easy to compute with direct substitution.
2. Surds, Roots, and Conjugates
The key algebraic identity for rationalisation is
###(\sqrt{a} – \sqrt{b})(\sqrt{a} + \sqrt{b}) = a – b.###
Here ###\sqrt{a} + \sqrt{b}### is the conjugate of ###\sqrt{a} – \sqrt{b}###, and vice versa. When we multiply an expression involving a difference of square roots by its conjugate, the square roots disappear and we are left with a simpler difference of the underlying expressions.
More generally, if we have a fraction of the form
###\frac{\sqrt{f(x)} – \sqrt{g(x)}}{h(x)},###
we can multiply numerator and denominator by ###\sqrt{f(x)} + \sqrt{g(x)}###. The numerator then becomes ##f(x) – g(x)##, which is often factorable or at least polynomial in nature. A similar idea works when the surd is in the denominator.
3. Rationalising the Numerator: Standard Pattern
Example 1 (Classic Square-Root Limit)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x}.###
Step 1: Direct substitution.
Substituting ##x = 0## gives ##\frac{0}{0}##, an indeterminate form. So we need to simplify.
Step 2: Multiply numerator and denominator by the conjugate.
The conjugate of ###\sqrt{1 + x} – 1### is ###\sqrt{1 + x} + 1###. Multiply by 1 in the form of a fraction:
###\frac{\sqrt{1 + x} – 1}{x} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}.###
Step 3: Use the identity to simplify the numerator.
###(\sqrt{1 + x} – 1)(\sqrt{1 + x} + 1) = (1 + x) – 1 = x.###
The whole expression becomes
###\frac{x}{x(\sqrt{1 + x} + 1)} = \frac{1}{\sqrt{1 + x} + 1}, \quad x \neq 0.###
Step 4: Take the limit of the simplified expression.
###\lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}.###
So
###\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x} = \frac{1}{2}.###
Example 2 (Square Root Approaching a Nonzero Point)
Evaluate
###\lim_{x \to 4} \frac{\sqrt{x} – 2}{x – 4}.###
Solution.
- Direct substitution gives ##\frac{2 – 2}{4 – 4} = \frac{0}{0}##.
- We rationalise the numerator using the conjugate ###\sqrt{x} + 2###:
###\frac{\sqrt{x} – 2}{x – 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}.###
The numerator simplifies to ##x – 4##, so we get
###\frac{x – 4}{(x – 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}, \quad x \neq 4.###
Now we can take the limit:
###\lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}.###
Example 3 (General Square-Root Difference)
Evaluate
###\lim_{x \to a} \frac{\sqrt{x} – \sqrt{a}}{x – a}, \quad a > 0.###
Solution.
Direct substitution gives ##\frac{0}{0}##. We rationalise the numerator:
###\frac{\sqrt{x} – \sqrt{a}}{x – a} \cdot \frac{\sqrt{x} + \sqrt{a}}{\sqrt{x} + \sqrt{a}}.###
The numerator becomes ##x – a##, so
###\frac{x – a}{(x – a)(\sqrt{x} + \sqrt{a})} = \frac{1}{\sqrt{x} + \sqrt{a}}, \quad x \neq a.###
Now let ##x \to a##:
###\lim_{x \to a} \frac{\sqrt{x} – \sqrt{a}}{x – a} = \frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}.###
This general result appears frequently in problems on limits using rationalisation in calculus.
Example 4 (Root of a Linear Expression)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{1 + 2x} – 1}{x}.###
Solution.
We rationalise the numerator using ###\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 the fraction simplifies to
###\frac{2x}{x(\sqrt{1 + 2x} + 1)} = \frac{2}{\sqrt{1 + 2x} + 1}, \quad x \neq 0.###
Now take the limit:
###\lim_{x \to 0} \frac{2}{\sqrt{1 + 2x} + 1} = \frac{2}{\sqrt{1} + 1} = \frac{2}{1 + 1} = 1.###
Example 5 (Root at a Nonzero Point)
Evaluate
###\lim_{x \to 9} \frac{\sqrt{x} – 3}{x – 9}.###
Solution.
We rationalise the numerator using ###\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.###
Now
###\lim_{x \to 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{3 + 3} = \frac{1}{6}.###
Example 6 (Two Surds in the Numerator)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{4 + x} – \sqrt{4 – x}}{x}.###
Solution.
We view the numerator as ##\sqrt{4 + x} – \sqrt{4 – x}## and rationalise it using the conjugate ###\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 the entire expression simplifies to
###\frac{2x}{x(\sqrt{4 + x} + \sqrt{4 – x})} = \frac{2}{\sqrt{4 + x} + \sqrt{4 – x}}, \quad x \neq 0.###
Now take the limit:
###\lim_{x \to 0} \frac{2}{\sqrt{4 + x} + \sqrt{4 – x}} = \frac{2}{\sqrt{4} + \sqrt{4}} = \frac{2}{2 + 2} = \frac{1}{2}.###
4. Rationalising the Denominator
Sometimes the surd appears in the denominator, and we may wish to rationalise the denominator instead.
Example 7 (Basic Denominator Rationalisation)
Evaluate
###\lim_{x \to 0} \frac{x}{\sqrt{1 + x} – 1}.###
Solution.
Direct substitution gives ##\frac{0}{0}##. We multiply numerator and denominator by the conjugate of the denominator, ###\sqrt{1 + x} + 1###:
###\frac{x}{\sqrt{1 + x} – 1} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}.###
The denominator becomes
###(\sqrt{1 + x} – 1)(\sqrt{1 + x} + 1) = (1 + x) – 1 = x.###
So the expression simplifies to
###\frac{x(\sqrt{1 + x} + 1)}{x} = \sqrt{1 + x} + 1, \quad x \neq 0.###
Now take the limit:
###\lim_{x \to 0} (\sqrt{1 + x} + 1) = \sqrt{1} + 1 = 2.###
Example 8 (Denominator with a Minus Sign)
Evaluate
###\lim_{x \to 0} \frac{x}{\sqrt{1 – x} – 1}.###
Solution.
Again, substitution gives ##\frac{0}{0}##. Multiply numerator and denominator 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.###
So the expression simplifies to
###\frac{x(\sqrt{1 – x} + 1)}{-x} = -(\sqrt{1 – x} + 1), \quad x \neq 0.###
Now
###\lim_{x \to 0} -(\sqrt{1 – x} + 1) = -(\sqrt{1} + 1) = -(1 + 1) = -2.###
Example 9 (Denominator with a Constant Inside the Root)
Evaluate
###\lim_{x \to 0} \frac{x}{\sqrt{4 + x} – 2}.###
Solution.
We rationalise the denominator using ###\sqrt{4 + x} + 2###:
###\frac{x}{\sqrt{4 + x} – 2} \cdot \frac{\sqrt{4 + x} + 2}{\sqrt{4 + x} + 2}.###
The denominator is
###(4 + x) – 4 = x.###
So we obtain
###\frac{x(\sqrt{4 + x} + 2)}{x} = \sqrt{4 + x} + 2, \quad x \neq 0.###
Now
###\lim_{x \to 0} (\sqrt{4 + x} + 2) = \sqrt{4} + 2 = 2 + 2 = 4.###
5. Mixed Patterns and More Involved Examples
Example 10 (Root of a Quadratic Over a Power of x)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{x^2 + 4} – 2}{x^2}.###
Solution.
Rationalise the numerator using ###\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.###
So the expression simplifies to
###\frac{x^2}{x^2(\sqrt{x^2 + 4} + 2)} = \frac{1}{\sqrt{x^2 + 4} + 2}, \quad x \neq 0.###
Now take the limit:
###\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 11 (Two Different Linear Expressions Inside Roots)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{2x + 3} – \sqrt{3}}{x}.###
Solution.
We rationalise the numerator using ###\sqrt{2x + 3} + \sqrt{3}###:
###\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 the expression simplifies to
###\frac{2x}{x(\sqrt{2x + 3} + \sqrt{3})} = \frac{2}{\sqrt{2x + 3} + \sqrt{3}}, \quad x \neq 0.###
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 12 (Symmetric Square Roots)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{a + x} – \sqrt{a – x}}{x}, \quad a > 0.###
Solution.
Rationalise the numerator using ###\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
###\frac{2x}{x(\sqrt{a + x} + \sqrt{a – x})} = \frac{2}{\sqrt{a + x} + \sqrt{a – x}}, \quad x \neq 0.###
Now take the limit:
###\lim_{x \to 0} \frac{2}{\sqrt{a + x} + \sqrt{a – x}} = \frac{2}{\sqrt{a} + \sqrt{a}} = \frac{2}{2\sqrt{a}} = \frac{1}{\sqrt{a}}.###
Example 13 (Surd Combined with a Polynomial)
Evaluate
###\lim_{x \to 0} \left( \frac{\sqrt{1 + x} – 1}{x} – \frac{1}{2} \right).###
This expression measures how well the approximation ##\sqrt{1 + x} \approx 1 + \frac{x}{2}## works near ##x = 0##.
Solution.
We already know from Example 1 that
###\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x} = \frac{1}{2}.###
Therefore,
###\lim_{x \to 0} \left( \frac{\sqrt{1 + x} – 1}{x} – \frac{1}{2} \right) = \frac{1}{2} – \frac{1}{2} = 0.###
So the difference tends to 0, confirming that the approximation is very accurate near ##x = 0##.
Example 14 (Derivative-Flavoured Limit with Roots)
Evaluate
###\lim_{h \to 0} \frac{\sqrt{x + h} – \sqrt{x}}{h}, \quad x > 0.###
Solution.
Rationalise the numerator using ###\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.###
So the fraction simplifies to
###\frac{h}{h(\sqrt{x + h} + \sqrt{x})} = \frac{1}{\sqrt{x + h} + \sqrt{x}}, \quad h \neq 0.###
Now take the limit as ##h \to 0##:
###\lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}.###
This is the derivative of ##\sqrt{x}## with respect to ##x##, obtained using limits using rationalisation in calculus.
Example 15 (Combination of Root and Polynomial, Another Style)
Evaluate
###\lim_{x \to 0} \frac{\sqrt{4 + x} – 2}{x}.###
Solution.
We rationalise the numerator using ###\sqrt{4 + x} + 2###:
###\frac{\sqrt{4 + x} – 2}{x} \cdot \frac{\sqrt{4 + x} + 2}{\sqrt{4 + x} + 2}.###
The numerator becomes
###(4 + x) – 4 = x.###
So we obtain
###\frac{x}{x(\sqrt{4 + x} + 2)} = \frac{1}{\sqrt{4 + x} + 2}, \quad x \neq 0.###
Now
###\lim_{x \to 0} \frac{1}{\sqrt{4 + x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}.###
6. Summary of the Method
In this reading, we deepened our understanding of limits using rationalisation in calculus. We saw that:
- Expressions involving square roots often lead to the indeterminate form ##\frac{0}{0}## when we attempt direct substitution.
- Multiplying by a suitable conjugate (for the numerator or denominator) converts a difference of square roots into a difference of simpler expressions.
- After rationalisation, a factor responsible for ##\frac{0}{0}## usually cancels, leaving a function whose limit is easy to compute.
- The same basic idea works for many patterns: single square roots, sums and differences of roots, roots of linear expressions, and roots of quadratics.
- This technique also appears naturally in derivative-style limits, such as the limits that define the derivatives of square-root functions.
With these examples and methods, we are now well prepared to tackle limits involving surds in more complex settings and to move on to trigonometric and exponential limits that appear throughout calculus.
7. References for Further Exploration
- The article on limits in calculus at Wikipedia provides a broad overview of limit techniques and examples.
- Expository notes and problem collections from established organisations such as the American Mathematical Society (AMS) and the National Institute of Standards and Technology (NIST) often include discussions of limits and related algebraic tools.
0 Comments