Physics and Mathematics
Limits Form 1 (Zero by Zero)
1. Concept Overview
A limit is said to be in Zero by Zero form when:
[\lim_{x \to a} \dfrac{f(x)}{g(x)}]
and on direct substitution [x = a]:
[f(a) = 0]
[g(a) = 0]
So the expression becomes:
[\dfrac{0}{0}]
This is called an indeterminate form.
2. Why (0/0) Is Called Indeterminate
The form [0/0] does not give any information about the limit.
Example:
- [\dfrac{x}{x}] → limit is [1]
- [\dfrac{x^2}{x}] → limit is [0]
- [\dfrac{x}{x^2}] → limit is [∞]
All give [0/0] at [x = 0], but results are different.
Therefore, we must simplify before evaluating the limit.
3. When Does (0/0) Appear?
(0/0) usually appears when:
- Polynomial expressions are involved
- Factors cancel after simplification
- Trigonometric functions approach zero
- Radicals vanish at the given point
4. General Method to Solve (0/0) Limits
Step-by-Step Strategy
- Substitute directly to confirm [0/0]
- Simplify the expression using:
- Factorisation
- Cancellation
- Rationalisation
- Standard limits (later)
- Remove the zero-causing factor
- Substitute again to find the limit
5. Simple Example
[\lim_{x \to 2} \dfrac{x^2 – 4}{x – 2}]
Step 1: Substitute [x = 2] → [0/0]
Step 2: Factorise numerator
[x^2 – 4 = (x – 2)(x + 2)]
Step 3: Cancel [x − 2]
Step 4: Substitute [x = 2]
Limit exists and is finite.
6. Important Points to Remember
- Never cancel before confirming (0/0)
- Cancellation is allowed only after factorisation
- The cancelled factor corresponds to the point where the function is undefined
- Limit may exist even if function value does not
7. Key Features
- (0/0) is the most common indeterminate form
- Requires algebraic manipulation
- Basis of continuity and differentiability
- Appears frequently in board examinations
8. Conceptual Questions with Solutions
1. What does the form (0/0) indicate?
It indicates that direct substitution fails and the limit value cannot be determined without simplification. It does NOT mean the limit is zero.
2. Why can’t we directly evaluate (0/0)?
Because division by zero is undefined, and both numerator and denominator becoming zero gives no clue about the limit value.
3. Does (0/0) always mean the limit exists?
No. After simplification, the limit may exist, be infinite, or not exist.
4. Why is factorisation important in (0/0) limits?
Factorisation helps isolate and remove the zero-causing factor, allowing evaluation of the remaining expression.
5. Can cancellation change the limit?
No. Proper cancellation does not change the limit, but cancelling incorrectly can lead to wrong answers.
6. Is (0/0) the same as undefined?
No. It means “cannot decide yet,” not “does not exist.”
7. Why do polynomial limits often give (0/0)?
Because polynomials are continuous, and factors often vanish at the point of approach.
8. Can trigonometric limits produce (0/0)?
Yes, especially when sine or tangent approaches zero.
9. What happens if simplification still gives (0/0)?
Further techniques such as standard limits are required.
10. Why is (0/0) common in continuity problems?
Because continuity checks involve limits at points where functions may be undefined.
11. Can a function be undefined but limit exists?
Yes. This is common in removable discontinuities.
12. Should LHL and RHL be checked for (0/0)?
Only if the simplified expression behaves differently on each side.
13. Is rationalisation useful in (0/0) limits?
Yes, especially when square roots are involved.
14. What is the biggest beginner mistake?
Cancelling terms without factorising.
15. Why must simplification be done before substitution?
Because substitution alone leads to an undefined expression.
9. FAQ / Common Misconceptions
1. (0/0) means the answer is zero.
False. It gives no information about the answer.
2. We can cancel zeros directly.
Incorrect. Zeros are results, not factors.
3. Every (0/0) limit exists.
False.
4. Simplification always gives a finite value.
Not necessarily.
5. Cancellation changes the function.
It changes the function value, not the limit.
6. Factorisation is optional.
No, it is essential.
7. (0/0) means discontinuity.
Not always.
8. Rationalisation is advanced.
It is basic and necessary.
9. Polynomial limits never fail.
They fail when division by zero occurs.
10. Limits and functions are unrelated.
They are closely connected.
10. Practice Questions with Step-by-Step Solutions
Question 1. Evaluate:
[\lim_{x \to 1} \dfrac{x^2 – 1}{x – 1}]
Step-by-Step Solution:
Step 1: Direct Substitution
Put [x = 1] in the given expression.
Numerator:
[x^2 − 1 = 1 − 1 = 0]
Denominator:
[x − 1 = 0]
So, the limit becomes [0/0], which is an indeterminate form.
Step 2: Factorisation
Factorise the numerator using the identity
[a^2 − b^2 = (a − b)(a + b)].
[x^2 − 1 = (x − 1)(x + 1)]
Step 3: Cancellation
Cancel the common factor [(x − 1)] from numerator and denominator.
The expression becomes:
[x + 1]
Step 4: Substitute Again
Now put [x = 1].
[x + 1 = 1 + 1 = 2]
Final Answer:
[\lim_{x \to 1} \dfrac{x^2 – 1}{x – 1} = 2]
Question 2. Evaluate:
[\lim_{x \to 2} \dfrac{x^2 – 4}{x – 2}]
Step-by-Step Solution:
Step 1: Direct Substitution
Numerator:
[x^2 − 4 = 4 − 4 = 0]
Denominator:
[x − 2 = 0]
So we get [0/0], which is indeterminate.
Step 2: Factorisation
[x^2 − 4 = (x − 2)(x + 2)]
Step 3: Cancellation
Cancel [(x − 2)].
Remaining expression:
[x + 2]
Step 4: Substitute [x = 2]
[x + 2 = 2 + 2 = 4]
Final Answer:
[\lim_{x \to 2} \dfrac{x^2 – 4}{x – 2} = 4]
Question 3. Evaluate:
[\lim_{x \to 0} \dfrac{x^2}{x}]
Step-by-Step Solution:
Step 1: Direct Substitution
Numerator:
[x^2 = 0]
Denominator:
[x = 0]
So we get [0/0].
Step 2: Simplification
[x^2 = x · x]
So the expression becomes:
[\dfrac{x · x}{x}]
Cancel [x].
Remaining expression:
[x]
Step 3: Substitute [x = 0]
[x = 0]
Final Answer:
[\lim_{x \to 0} \dfrac{x^2}{x} = 0]
Question 4. Evaluate:
[\lim_{x \to 3} \dfrac{x^2 – 9}{x – 3}]
Step-by-Step Solution:
Step 1: Direct Substitution
[x^2 − 9 = 9 − 9 = 0]
[x − 3 = 0]
So the form is [0/0].
Step 2: Factorisation
[x^2 − 9 = (x − 3)(x + 3)]
Step 3: Cancellation
Cancel [(x − 3)].
Remaining expression:
[x + 3]
Step 4: Substitute [x = 3]
[x + 3 = 6]
Final Answer:
[\lim_{x \to 3} \dfrac{x^2 – 9}{x – 3} = 6]
Question 5. 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 we get [0/0].
Step 2: Rationalisation
Multiply numerator and denominator by
[\sqrt{x} + 2].
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Step 3: Simplify
Numerator:
[(\sqrt{x})^2 − 4 = x − 4]
So expression becomes:
[\dfrac{x − 4}{(x − 4)(\sqrt{x} + 2)}]
Cancel [(x − 4)].
Remaining expression:
[\dfrac{1}{\sqrt{x} + 2}]
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 6. Evaluate:
[\lim_{x \to 0} \dfrac{x}{\sqrt{x^2}}]
Step-by-Step Solution:
Step 1: Rewrite the Denominator
[\sqrt{x^2} = |x|]
So the expression becomes:
[\dfrac{x}{|x|}]
Step 2: Left-Hand Limit (LHL)
For [x < 0], [|x| = −x]. So, [\dfrac{x}{|x|} = \dfrac{x}{−x} = −1] Step 3: Right-Hand Limit (RHL) For [x > 0], [|x| = x].
So,
[\dfrac{x}{|x|} = 1]
Final Conclusion:
LHL ≠ RHL
So the limit does not exist.
Final Answer:
Limit does not exist.
Question 7. Evaluate:
[\lim_{x \to 1} \dfrac{x^3 – 1}{x – 1}]
Step-by-Step Solution:
Step 1: Direct Substitution
[x^3 − 1 = 0]
[x − 1 = 0]
So we get [0/0].
Step 2: Factorisation
Use identity:
[a^3 − b^3 = (a − b)(a^2 + ab + b^2)]
[x^3 − 1 = (x − 1)(x^2 + x + 1)]
Step 3: Cancellation
Cancel [(x − 1)].
Remaining expression:
[x^2 + x + 1]
Step 4: Substitute [x = 1]
[1 + 1 + 1 = 3]
Final Answer:
[\lim_{x \to 1} \dfrac{x^3 – 1}{x – 1} = 3]
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