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Limits and Algebra of Limits

🎓 Class 11 Mathematics CBSE Theory Ch 12 — Limits and Derivatives ⏱ ~30 min
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This MCQ module is based on: Limits and Algebra of Limits

This mathematics assessment will be based on: Limits and Algebra of Limits
Targeting Class 11 level in Calculus, with Advanced difficulty.

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12.3 Limits

The idea of instantaneous velocity used the statement "as \(h\to 0\)". This is a limit?. In this section we formalise the notion of the limit of a function at a point, look at left-hand and right-hand limits, and develop the algebra of limits.

Limit of a Function (Working Definition)
We say that \(\displaystyle \lim_{x\to a} f(x)=L\) if the values \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(a\) (but not equal to \(a\)).

Left Hand and Right Hand Limits

If we approach \(a\) through values smaller than \(a\), the limit is called the left-hand limit (LHL), written \(\displaystyle \lim_{x\to a^-}f(x)\). If we approach through values larger than \(a\), it is the right-hand limit (RHL), \(\displaystyle \lim_{x\to a^+}f(x)\).

Existence of Limit
\(\lim_{x\to a}f(x)\) exists if and only if LHL = RHL. Their common value is then the limit. If LHL ≠ RHL, the limit does not exist.

Example 2

Find \(\displaystyle \lim_{x\to 1} f(x)\), where \(f(x)=\begin{cases} x+10,&x\le 1\\ 2x+3,&x>1\end{cases}\).

LHL: \(\lim_{x\to 1^-}(x+10)=11\). RHL: \(\lim_{x\to 1^+}(2x+3)=5\). Since LHL ≠ RHL, the limit does not exist.

Example 3 — Numerical Approach

Consider \(f(x)=x^2\) near \(x=1\):

x0.90.990.9991.0011.011.1
f(x)0.810.98010.9980011.0020011.02011.21

From both sides, \(f(x)\to 1\). So \(\lim_{x\to 1} x^2 = 1\).

x y 1 1 L=1 x→1⁻ x→1⁺
Fig 12.2: As \(x\to 1\) from either side, \(x^2\to 1\).

12.3.1 Algebra of Limits

Let \(f\) and \(g\) be two functions with \(\displaystyle \lim_{x\to a} f(x)=L\) and \(\displaystyle \lim_{x\to a} g(x)=M\). Then:

Limit Laws
  1. Sum: \(\lim_{x\to a}[f(x)+g(x)]=L+M\)
  2. Difference: \(\lim_{x\to a}[f(x)-g(x)]=L-M\)
  3. Product: \(\lim_{x\to a}[f(x)\cdot g(x)]=L\cdot M\)
  4. Scalar: \(\lim_{x\to a}\bigl[k\,f(x)\bigr]=kL\)
  5. Quotient: \(\lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{L}{M}\) provided \(M\ne 0\)

Limits of Polynomials and Rational Functions

If \(f(x)=a_n x^n+a_{n-1}x^{n-1}+\dots +a_0\) is any polynomial, then \(\lim_{x\to a} f(x)=f(a)\) — i.e., polynomial limits are found simply by substitution.

For a rational function \(f(x)=\dfrac{p(x)}{q(x)}\) with \(q(a)\ne 0\), \(\lim_{x\to a} f(x)=\dfrac{p(a)}{q(a)}\). If \(q(a)=0\) but \(p(a)=0\), we factor and cancel the common factor before substituting.

Example 4

Find \(\displaystyle \lim_{x\to 1}[x^3-x^2+1]\).

Polynomial ⟹ substitute: \(1-1+1=1\).

Example 5

Find \(\displaystyle \lim_{x\to 2}\frac{x^3-4x^2+4x}{x^2-4}\).

At \(x=2\): numerator \(=8-16+8=0\); denominator \(=4-4=0\). \(0/0\) form — factor.
Numerator \(=x(x^2-4x+4)=x(x-2)^2\). Denominator \(=(x-2)(x+2)\).
Cancel \((x-2)\): limit \(=\displaystyle\lim_{x\to 2}\frac{x(x-2)}{x+2}=\frac{2\cdot 0}{4}=0\).

Example 6 — Standard Limit

Prove \(\displaystyle \lim_{x\to a}\frac{x^n-a^n}{x-a}=n\,a^{n-1}\) for any positive integer \(n\).

Factor: \(x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\dots+x a^{n-2}+a^{n-1})\).
Cancel \((x-a)\): \(\displaystyle\lim_{x\to a}(x^{n-1}+x^{n-2}a+\dots+a^{n-1})\).
There are \(n\) terms; as \(x\to a\), each term tends to \(a^{n-1}\). Hence the limit is \(n\,a^{n-1}\).

Example 7

Evaluate \(\displaystyle \lim_{x\to 1}\frac{x^{15}-1}{x^{10}-1}\).

Divide numerator and denominator by \((x-1)\) using Example 6: \(\displaystyle\lim_{x\to 1}\frac{x^{15}-1}{x-1}=15\); \(\displaystyle\lim_{x\to 1}\frac{x^{10}-1}{x-1}=10\).
Therefore \(\dfrac{15}{10}=\dfrac{3}{2}\).

Example 8

Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sqrt{1+x}-1}{x}\).

Rationalise: multiply top and bottom by \(\sqrt{1+x}+1\): \(\displaystyle\frac{(1+x)-1}{x(\sqrt{1+x}+1)}=\frac{1}{\sqrt{1+x}+1}\).
As \(x\to 0\), the expression \(\to\dfrac{1}{2}\).
Activity: Numerical Exploration of a Limit
L3 Apply
Materials: Calculator, table-sheet.
Predict: What value will \(\dfrac{x^2-9}{x-3}\) approach as \(x\to 3\)?
  1. Evaluate \(\dfrac{x^2-9}{x-3}\) for \(x=2.9, 2.99, 2.999\) and \(x=3.1, 3.01, 3.001\).
  2. Tabulate your values to 4 decimal places.
  3. Conjecture the limit.
  4. Verify algebraically by factoring and cancelling.
  5. Discuss: why does direct substitution fail, yet the limit still exists?
Values approach 6 from both sides. Algebraically, \(\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3\to 6\) as \(x\to 3\). Direct substitution gives \(\tfrac{0}{0}\) because \(x=3\) is a removable discontinuity — the cancelled factor is the one vanishing in both numerator and denominator.

Exercise 12.1 (Selected Problems)

Q1. Evaluate \(\displaystyle \lim_{x\to 3}(x+3)\).
Direct substitution: \(3+3=6\).
Q2. Evaluate \(\displaystyle \lim_{x\to \pi}\left(x-\tfrac{22}{7}\right)\).
\(\pi-\tfrac{22}{7}\).
Q3. Evaluate \(\displaystyle \lim_{r\to 1}\pi r^2\).
\(\pi\).
Q4. Evaluate \(\displaystyle \lim_{x\to 4}\dfrac{4x+3}{x-2}\).
Rational, denominator non-zero: \(\dfrac{16+3}{4-2}=\dfrac{19}{2}\).
Q5. Evaluate \(\displaystyle \lim_{x\to -1}\dfrac{x^{10}+x^5+1}{x-1}\).
\(\dfrac{1-1+1}{-1-1}=\dfrac{1}{-2}=-\dfrac{1}{2}\).
Q6. Evaluate \(\displaystyle \lim_{x\to 0}\dfrac{(x+1)^5-1}{x}\).
Let \(u=x+1\), so \(u\to 1\). Limit \(=\displaystyle\lim_{u\to 1}\dfrac{u^5-1}{u-1}=5\cdot 1^4=5\) (Example 6).
Q7. Evaluate \(\displaystyle \lim_{x\to 1}\dfrac{x^{15}-1}{x^{10}-1}\).
\(\dfrac{15}{10}=\dfrac{3}{2}\) (see Example 7).
Q8. Evaluate \(\displaystyle \lim_{z\to 1}\dfrac{z^{1/3}-1}{z^{1/6}-1}\).
Let \(u=z^{1/6}\), so \(u\to 1\) and \(z^{1/3}=u^2\). Limit \(=\displaystyle\lim_{u\to 1}\dfrac{u^2-1}{u-1}=\lim_{u\to 1}(u+1)=2\).

Competency-Based Questions

Scenario: A student computes \(\displaystyle \lim_{x\to 2}\dfrac{x^2-4}{x-2}\) by substituting \(x=2\) and gets \(\tfrac{0}{0}\). She concludes "the limit does not exist" and stops.
Q1. The correct value of the limit is:
L3 Apply
  • (a) Does not exist
  • (b) 0
  • (c) 2
  • (d) 4
Answer: (d) 4. Factor: \(\dfrac{(x-2)(x+2)}{x-2}=x+2\to 4\).
Q2. Analyse: why is "\(0/0\)" not the same as "the limit does not exist"?
L4 Analyse
Answer: The form \(0/0\) is called an indeterminate form: it tells us we cannot conclude the limit by substitution alone. The actual limit could be any real number, \(\pm\infty\), or may not exist — we must simplify (factor, rationalise, or use a standard limit) before evaluating.
Q3. Evaluate: for a piecewise function \(f\) defined on a small deleted neighbourhood of \(a\), the limit \(\lim_{x\to a} f(x)\) exists iff LHL = RHL. Use this to decide whether \(\lim_{x\to 0}\frac{|x|}{x}\) exists.
L5 Evaluate
Answer: For \(x>0\), \(|x|/x=1\); for \(x<0\), \(|x|/x=-1\). RHL = 1, LHL = −1. They are unequal, so the limit does not exist.
Q4. Create a rational function whose limit at \(x=5\) is \(\tfrac{1}{4}\) but where direct substitution gives \(\tfrac{0}{0}\).
L6 Create
Sample: \(f(x)=\dfrac{(x-5)(x-1)}{(x-5)(4x-16)}=\dfrac{x-1}{4x-16}\). At \(x=5\): direct gives \(0/0\); after cancelling, \(\dfrac{4}{4}=1\). Adjust coefficients to land at \(\tfrac{1}{4}\): use \(\dfrac{(x-5)(x-1)}{(x-5)(4)(x-1)+\text{stuff}}\). Cleaner example: \(f(x)=\dfrac{(x-5)}{4(x-5)(x-1)}\cdot (x-1)=\dfrac{1}{4}\) (constant after cancellation) — valid. Many answers.

Assertion–Reason Questions

Assertion (A): \(\displaystyle \lim_{x\to 2}\frac{x^2-4}{x-2}=4\).
Reason (R): \(x^2-4=(x-2)(x+2)\) allows cancellation of \(x-2\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R justifies the simplification that gives A.
Assertion (A): \(\displaystyle \lim_{x\to 0}\frac{|x|}{x}\) does not exist.
Reason (R): \(\frac{|x|}{x}\) is undefined at \(x=0\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (b). Both statements are true, but R is not the correct reason — a function may be undefined at a point and still have a limit there (e.g., \(\tfrac{x^2-4}{x-2}\) at \(x=2\)). The actual reason is LHL ≠ RHL.
Assertion (A): \(\displaystyle \lim_{x\to a}\frac{x^n-a^n}{x-a}=n\,a^{n-1}\).
Reason (R): \(x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\dots+a^{n-1})\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). The factorisation in R, followed by cancellation and substitution, yields \(n\,a^{n-1}\). R correctly explains A.

Frequently Asked Questions

When does a limit exist?
A limit exists at x = a if both one-sided limits (left and right) exist and are equal. If they differ, the limit does not exist.
What is the algebra of limits?
If limit f(x) = L and limit g(x) = M (as x approaches a), then limit (f plus-or-minus g) = L plus-or-minus M, limit (f g) = L M, and limit (f/g) = L/M when M is not 0.
What is limit as x approaches a of (x^n - a^n)/(x - a)?
It equals n times a^(n-1). This is a standard NCERT limit result used widely.
What is the left-hand limit?
Left-hand limit (LHL) is the value f(x) approaches as x tends to a from values less than a. Notation: limit as x approaches a minus of f(x).
Why can direct substitution sometimes fail?
Direct substitution fails when it produces an indeterminate form like 0/0 or infinity/infinity - the limit may still exist but needs simplification, factorisation or standard results.
Are limits used in Class 12?
Yes. Limits are the foundation for continuity, differentiation and integration in Class 12 Calculus.
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