Limits: one-sided behaviour and standard forms

Goals and prerequisites

Use algebra and elementary trigonometry to evaluate real single-variable limits. All angles are in radians. Distinguish a limit from a function value and distinguish a finite limit from unbounded behaviour.

1. What approaches what?

The statement lim(x→a) f(x)=L says that f(x) becomes arbitrarily close to L whenever x is sufficiently close to a but x≠a, within the domain. Formally, for every ε>0 there is δ>0 such that 0<|x−a|<δ implies |f(x)−L|<ε. The point a must be an accumulation point of the domain. The value f(a) can be different or undefined.

A two-sided limit at an interior point exists exactly when the left and right limits exist and agree as finite real numbers. For f(x)=|x|/x, the left limit at zero is −1 and the right limit is 1. Thus no two-sided limit exists. An infinite limit describes divergence: 1/x²→+∞ as x→0, but this is not a finite real limit. For 1/x, the sides diverge with opposite signs.

2. Algebra before advanced rules

Limits distribute over sums and products when the component limits are finite. A quotient rule additionally requires a nonzero denominator limit. The symbol 0/0 is an indeterminate form, not an answer.

Example: lim(x→2)(x²−4)/(x−2). For x≠2, factor and cancel to obtain x+2. The limit is 4, although the original quotient is undefined at 2. Cancellation changes the value at a missing point, not the punctured-neighbourhood limit.

Example: lim(x→0)(√(1+x)−1)/x. The real domain near zero has x≥−1. Multiply numerator and denominator by √(1+x)+1 to obtain 1/(√(1+x)+1), so the limit is 1/2. Rationalization exposes the cancellation; substituting into 0/0 does not.

3. Squeezing and standard small arguments

If g(x)≤f(x)≤h(x) near a and both bounding limits equal L, the squeeze theorem gives lim f=L. Since −x²≤x²sin(1/x)≤x² for x≠0, its limit at zero is 0 despite the oscillation of sin(1/x). Bounded oscillation multiplied by something tending to zero can converge; sin(1/x) alone does not.

Standard limits: sin u/u→1, (1−cos u)/u²→1/2, (eᵘ−1)/u→1 and ln(1+u)/u→1 as u→0 (with u>−1 for the logarithm). For example, sin(3x)/(2x)=(3/2)[sin(3x)/(3x)]→3/2. Also (1−cos 2x)/x²=4[(1−cos 2x)/(2x)²]→2. Keep the scaling factors.

For positive bases near 1, (1+x)^(1/x)→e as x→0, with 1+x>0. One safe way to study powers is to take logarithms on a positive domain, find the logarithm's limit, then exponentiate.

4. Infinity and growth

For a rational function as x→+∞ or −∞, divide numerator and denominator by the highest denominator power. For (3x²−x+1)/(2x²+5), the limit is 3/2 at either end. If numerator degree is smaller the limit is zero; if larger, examine the leading power and its sign rather than merely writing infinity.

The hierarchy ln x ≪ xᵖ ≪ eˣ for fixed p>0 and x→+∞ means each slower/faster ratio tends to zero. It is a statement about this direction and domain, not a comparison valid for every x.

5. Conditional tool and checking routine

L'Hôpital's rule is a conditional tool for 0/0 or ∞/∞ forms. On the relevant punctured one-sided interval, require differentiability of numerator and denominator, nonzero denominator derivative, and existence of the derivative-quotient limit (possibly infinite); then the original quotient has that limit. It is not a license to differentiate arbitrary quotients. For (eˣ−1)/x, the derivative quotient is eˣ/1→1. For x→0, (1+x)/(2+x) already has limit 1/2; differentiating would wrongly produce 1.

Checklist: identify domain and approach side, try substitution, classify any indeterminate form, simplify or squeeze, and check both sides when required. No graph alone proves a limit.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

CL01. Find lim(x→2)(x²−4)/(x−2).

A. 0 B. 2 C. 4 D. Does not exist / मौजूद नहीं

Answer: C

A: Zero over zero is not zero.

B: After cancellation the expression is x+2, not x.

C: For x≠2 it equals x+2, whose limit is 4.

D: A missing value at 2 does not prevent a limit.

CL02. What is the two-sided limit of |x|/x at zero?

A. 1 B. −1 C. 0 D. Does not exist / मौजूद नहीं

Answer: D

A: Only the right limit is 1.

B: Only the left limit is −1.

C: Averaging side limits is invalid.

D: The side limits −1 and 1 disagree.

CL03. Find lim(x→0) sin(3x)/(2x), in radians.

A. 1 B. 2/3 C. 3/2 D. 3

Answer: C

A: The scaling factor is 3/2.

B: This reverses the scaling ratio.

C: Rewrite as (3/2)sin(3x)/(3x).

D: The denominator factor 2 remains.

CL04. Find lim(x→0) x² sin(1/x).

A. 0 B. 1 C. +∞ D. Does not exist / मौजूद नहीं

Answer: A

A: Its absolute value is at most x²→0.

B: The shrinking bound forces zero.

C: The function is bounded near zero.

D: Oscillation is squeezed by x².

CL05. Find lim(x→+∞)(3x²−x+1)/(2x²+5).

A. 0 B. 3/2 C. +∞ D. 1

Answer: B

A: Equal degrees do not give zero here.

B: Divide by x²; lower-order terms vanish.

C: The leading powers cancel.

D: The leading coefficients differ.

CL06. Find lim(x→0)(√(1+x)−1)/x.

A. 1 B. 0 C. 2 D. 1/2

Answer: D

A: Rationalization gives denominator tending to 2.

B: The form 0/0 is indeterminate.

C: This takes the reciprocal incorrectly.

D: The expression equals 1/(√(1+x)+1) for x≠0.

Source and scope

Official Limits heading; original single-variable foundations and conditional supporting tools.

Official GATE 2026 CS syllabus (verified 2026-10-14).