Physics and Mathematics
Limits Rationalisation Method
1. Concept Overview
The rationalisation method is used to solve limits when square roots cause the indeterminate form (0/0).
It involves removing the square root from the numerator or denominator by multiplying with its conjugate.
2. What Is Rationalisation?
For expressions of the form:
- [\sqrt{a} − \sqrt{b}]
- [\sqrt{x} − a]
- [\sqrt{f(x)} − \sqrt{g(x)}]
We multiply by the conjugate:
- [\sqrt{a} + \sqrt{b}]
- [\sqrt{x} + a]
This uses the identity:
[(\sqrt{a} − \sqrt{b})(\sqrt{a} + \sqrt{b}) = a − b]
3. When Is Rationalisation Required?
Rationalisation is required when:
- Direct substitution gives [0/0]
- Square roots are present
- Factorisation is not possible
- Cancelling is not possible directly
4. General Method (Step-by-Step)
To evaluate
[\lim_{x \to a} f(x)]
Steps:
- Substitute [x = a]
- If result is [0/0], proceed further
- Multiply numerator and denominator by the conjugate
- Simplify using identities
- Cancel common factors
- Substitute again to find the limit
5. Simple Illustrative Example
[\lim_{x \to 4} \dfrac{\sqrt{x} − 2}{x − 4}]
- Substitute [x = 4] → [0/0]
- Multiply numerator and denominator by conjugate [\sqrt{x} + 2]
- Simplify numerator → [x − 4]
- Cancel [x − 4]
- Substitute again
Limit exists and is finite.
6. Key Features to Remember
- Used mainly for irrational expressions
- Based on difference of squares
- Removes square roots causing zero
- Very common in board exams
7. Conceptual Questions with Detailed Solutions (15)
1. Why is rationalisation needed in limits?
Rationalisation removes square roots that prevent simplification. Without it, cancellation is not possible and the indeterminate form [0/0] remains unresolved.
2. Can rationalisation be used if no square root is present?
No. Rationalisation is meaningful only when irrational expressions involving roots are present.
3. Why do square roots often cause (0/0)?
Because when the variable approaches a value that makes the square root equal to a constant, subtraction often leads to zero.
4. What is a conjugate?
The conjugate is formed by changing the sign between two terms, such as [\sqrt{x} − 2] becoming [\sqrt{x} + 2].
5. Why must we multiply numerator and denominator by the same expression?
To keep the value of the expression unchanged while enabling simplification.
6. Does rationalisation change the limit value?
No. It only simplifies the expression.
7. Is rationalisation compulsory whenever roots appear?
Only when roots lead to an indeterminate form.
8. Can we rationalise the denominator instead of numerator?
Yes, depending on where the root appears.
9. Why is factorisation not sufficient in such problems?
Because square roots do not factorise like polynomials.
10. What identity makes rationalisation possible?
The identity [(a − b)(a + b) = a^2 − b^2].
11. Can rationalisation lead to infinite limits?
Yes, if cancellation does not occur completely.
12. Why is this method important before continuity?
Because many continuity problems involve roots.
13. What is the most common mistake in rationalisation?
Forgetting to multiply both numerator and denominator.
14. Can rationalisation be combined with factorisation?
Yes, often both are required.
15. How do we know rationalisation is complete?
When the root causing zero is eliminated.
8. FAQ / Common Misconceptions (10)
1. Rationalisation is optional.
No. It is necessary when roots cause [0/0].
2. We can rationalise only the numerator.
Both numerator and denominator may be rationalised.
3. Rationalisation works only once.
Sometimes it must be applied repeatedly.
4. Roots cancel automatically.
They do not.
5. Rationalisation changes the limit.
False.
6. It is an advanced technique.
It is basic and essential.
7. Factorisation is better than rationalisation.
Both are complementary.
8. Square roots always give limits.
Only after simplification.
9. Rationalisation removes limits.
It removes indeterminacy, not limits.
10. Limits with roots are rare.
They are very common in exams.
9. Practice Questions with Step-by-Step Solutions
Question 1. Evaluate:
[\lim_{x \to 4} \dfrac{\sqrt{x} − 2}{x − 4}]
Step-by-Step Solution:
Step 1: Direct Substitution
[\sqrt{4} − 2 = 0]
[4 − 4 = 0]
So the form is [0/0].
Step 2: Multiply by Conjugate
Multiply numerator and denominator by [\sqrt{x} + 2].
[\dfrac{(\sqrt{x} − 2)(\sqrt{x} + 2)}{(x − 4)(\sqrt{x} + 2)}]
Step 3: Simplify
Numerator becomes:
[x − 4]
Expression becomes:
[\dfrac{x − 4}{(x − 4)(\sqrt{x} + 2)}]
Cancel [x − 4].
Step 4: Substitute [x = 4]
[\dfrac{1}{2 + 2} = \dfrac{1}{4}]
Final Answer:
[\lim_{x \to 4} \dfrac{\sqrt{x} − 2}{x − 4} = \dfrac{1}{4}]
Question 2. Evaluate:
[\lim_{x \to 9} \dfrac{\sqrt{x} − 3}{x − 9}]
Step-by-Step Solution:
Substitute → [0/0]
Multiply by [\sqrt{x} + 3]
Simplify numerator → [x − 9]
Cancel common factor
Substitute [x = 9]
Final Answer:
[\dfrac{1}{6}]
Question 3. Evaluate:
[\lim_{x \to 4} \dfrac{x − 4}{\sqrt{x} − 2}]
Step-by-Step Solution:
Step 1: Substitute [x = 4] directly
Numerator:
[4 − 4 = 0]
Denominator:
[\sqrt{4} − 2 = 2 − 2 = 0]
So, the limit is of the indeterminate form [0/0].
Step 2: Identify the cause of indeterminacy
The denominator contains a square root [\sqrt{x} − 2], which becomes zero.
Hence, we apply rationalisation.
Step 3: Multiply numerator and denominator by the conjugate
Conjugate of [\sqrt{x} − 2] is [\sqrt{x} + 2].
$ \displaystyle \frac{{x-4}}{{\sqrt{x}-2}}\times \frac{{\sqrt{x}+2}}{{\sqrt{x}+2}}$
Step 4: Simplify the denominator
Using identity [(a − b)(a + b) = a^2 − b^2]:
[(\sqrt{x})^2 − 2^2 = x − 4]
So the expression becomes:
[\dfrac{(x − 4)(\sqrt{x} + 2)}{x − 4}]
Step 5: Cancel common factor
Cancel [(x − 4)] from numerator and denominator:
[\sqrt{x} + 2]
Step 6: Substitute [x = 4]
[\sqrt{4} + 2 = 2 + 2 = 4]
Final Answer:
[\lim_{x \to 4} \dfrac{x − 4}{\sqrt{x} − 2} = 4]
Question 4. Evaluate:
[\lim_{x \to 1} \dfrac{\sqrt{x + 3} − 2}{x − 1}]
Step-by-Step Solution:
Step 1: Direct substitution
Numerator:
[\sqrt{1 + 3} − 2 = 2 − 2 = 0]
Denominator:
[1 − 1 = 0]
Indeterminate form [0/0].
Step 2: Rationalise the numerator
Multiply numerator and denominator by conjugate [\sqrt{x + 3} + 2]:
$ \displaystyle \frac{{\sqrt{{x+3}}-2}}{{x-1}}\times \frac{{\sqrt{{x+3}}+2}}{{\sqrt{{x+3}}+2}}$
Step 3: Simplify numerator
[(\sqrt{x + 3})^2 − 2^2 = x + 3 − 4 = x − 1]
So expression becomes:
[\dfrac{x − 1}{(x − 1)(\sqrt{x + 3} + 2)}]
Step 4: Cancel common factor
Cancel [(x − 1)]:
[\dfrac{1}{\sqrt{x + 3} + 2}]
Step 5: Substitute [x = 1]
[\dfrac{1}{2 + 2} = \dfrac{1}{4}]
Final Answer:
[\dfrac{1}{4}]
Question 5. Evaluate:
[\lim_{x \to 0} \dfrac{\sqrt{1 + x} − 1}{x}]
Step-by-Step Solution:
Step 1: Substitute [x = 0]
Numerator:
[\sqrt{1} − 1 = 0]
Denominator:
[0]
So the limit is [0/0].
Step 2: Rationalise the numerator
Multiply by conjugate [\sqrt{1 + x} + 1]:
$ \displaystyle \frac{{\sqrt{{1+x}}-1}}{x}\times \frac{{\sqrt{{1+x}}+1}}{{\sqrt{{1+x}}+1}}$
Step 3: Simplify numerator
[(1 + x) − 1 = x]
So expression becomes:
[\dfrac{x}{x(\sqrt{1 + x} + 1)}]
Step 4: Cancel [x]
[\dfrac{1}{\sqrt{1 + x} + 1}]
Step 5: Substitute [x = 0]
[\dfrac{1}{1 + 1} = \dfrac{1}{2}]
Final Answer:
[\dfrac{1}{2}]
Question 6. Evaluate:
[\lim_{x \to 0} \dfrac{x}{\sqrt{1 + x} − 1}]
Step-by-Step Solution:
Step 1: Substitute [x = 0]
Numerator: [0]
Denominator: [1 − 1 = 0]
So the limit is [0/0].
Step 2: Rationalise the denominator
Multiply by conjugate [\sqrt{1 + x} + 1]:
$\displaystyle \frac{x}{{\sqrt{{1+x}}-1}}\times \frac{{\sqrt{{1+x}}+1}}{{\sqrt{{1+x}}+1}}$
Step 3: Simplify denominator
[(1 + x) − 1 = x]
Expression becomes:
[\dfrac{x(\sqrt{1 + x} + 1)}{x}]
Step 4: Cancel [x]
[\sqrt{1 + x} + 1]
Step 5: Substitute [x = 0]
[1 + 1 = 2]
Final Answer:
[2]
Question 7. Evaluate:
[\lim_{x \to 0} \dfrac{\sqrt{a^2 + x} − a}{x}]
Step-by-Step Solution:
Step 1: Substitute [x = 0]
[\sqrt{a^2} − a = a − a = 0]
So we get [0/0].
Step 2: Rationalise numerator
Multiply by conjugate [\sqrt{a^2 + x} + a]:
$ \displaystyle \frac{{\sqrt{{{{a}^{2}}+x}}-a}}{x}\times \frac{{\sqrt{{{{a}^{2}}+x}}+a}}{{\sqrt{{{{a}^{2}}+x}}+a}}$​
Step 3: Simplify numerator
[(a^2 + x) − a^2 = x]
So expression becomes:
[\dfrac{x}{x(\sqrt{a^2 + x} + a)}]
Step 4: Cancel [x]
[\dfrac{1}{\sqrt{a^2 + x} + a}]
Step 5: Substitute [x = 0]
[\dfrac{1}{a + a} = \dfrac{1}{2a}]
Final Answer:
[\dfrac{1}{2a}]
Question 8. Evaluate:
[\lim_{x \to a} \dfrac{\sqrt{x} − \sqrt{a}}{x − a}]
Step-by-Step Solution:
Step 1: Substitute [x = a]
[\sqrt{a} − \sqrt{a} = 0]
[a − a = 0]
Indeterminate form [0/0].
Step 2: Rationalise numerator
Multiply by conjugate [\sqrt{x} + \sqrt{a}]:
$ \displaystyle \frac{{\sqrt{x}-\sqrt{a}}}{{x-a}}\times \frac{{\sqrt{x}+\sqrt{a}}}{{\sqrt{x}+\sqrt{a}}}$​
Step 3: Simplify numerator
[x − a]
Expression becomes:
[\dfrac{x − a}{(x − a)(\sqrt{x} + \sqrt{a})}]
Step 4: Cancel [(x − a)]
[\dfrac{1}{\sqrt{x} + \sqrt{a}}]
Step 5: Substitute [x = a]
[\dfrac{1}{2\sqrt{a}}]
Final Answer:
[\dfrac{1}{2\sqrt{a}}]
Question 9. Evaluate:
[\lim_{x \to 0} \dfrac{\sqrt{1 + 2x} − 1}{x}]
Step-by-Step Solution:
Step 1: Substitute [x = 0] → [0/0]
Step 2: Rationalise numerator
Multiply by conjugate [\sqrt{1 + 2x} + 1].
Step 3: Simplify numerator
[(1 + 2x) − 1 = 2x]
Expression becomes:
[\dfrac{2x}{x(\sqrt{1 + 2x} + 1)}]
Step 4: Cancel [x]
[\dfrac{2}{\sqrt{1 + 2x} + 1}]
Step 5: Substitute [x = 0]
[\dfrac{2}{1 + 1} = 1]
Final Answer:
[1]
Question 10. Evaluate:
[\lim_{x \to 0} \dfrac{\sqrt{1 + x} − \sqrt{1 − x}}{x}]
Step-by-Step Solution:
Step 1: Substitute [x = 0]
Numerator: [1 − 1 = 0]
Denominator: [0]
So the form is [0/0].
Step 2: Rationalise numerator
Multiply by conjugate [\sqrt{1 + x} + \sqrt{1 − x}]:
Step 3: Simplify numerator
[(1 + x) − (1 − x) = 2x]
Expression becomes:
[\dfrac{2x}{x(\sqrt{1 + x} + \sqrt{1 − x})}]
Step 4: Cancel [x]
[\dfrac{2}{\sqrt{1 + x} + \sqrt{1 − x}}]
Step 5: Substitute [x = 0]
[\dfrac{2}{1 + 1} = 1]
Final Answer:
[1]
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