Difficulty: Introductory | Prerequisites: Algebra (polynomial evaluation, piecewise functions, rationalisation)
Limits describe what value a function approaches as the input gets close to some number, even if the function never reaches that value. This recitation covers direct substitution for polynomials, the special limit sin(x)/x = 1, and how to handle piecewise functions where the two sides may disagree. When the left-hand and right-hand limits differ, the overall limit does not exist (DNE).
Limit
The value that f(x) approaches as x approaches a particular number c, written lim(x→c) f(x) = L. In simple terms, this means: "what y-value is the function heading towards as x gets close to c, whether or not it ever arrives."
One-sided limit
The value f(x) approaches from only the left (x→c⁻) or only the right (x→c⁺). Think of it as checking the function's behaviour from each direction independently, like approaching a junction from either road.
Piecewise function
A function defined by different rules on different intervals, e.g. f(x) = {3x + 7 if x ≠ 2, 5 if x = 2}. In simple terms, the function switches its formula depending on where x falls.
Limit Does Not Exist (DNE)
A limit fails to exist when the left-hand and right-hand limits are not equal, or when the function oscillates or blows up without settling on a single value. Think of it as: the two sides of the function cannot agree on where to meet.
Direct substitution
For continuous functions (most polynomials, rationals away from holes), you find the limit by simply plugging in the value of x. In simple terms: if nothing weird happens at that point, just substitute and compute.
Continuity at a point
A function is continuous at x = c when three things hold: f(c) is defined, lim(x→c) f(x) exists, and lim(x→c) f(x) = f(c). Think of it as: you can draw through that point without lifting your pen.
When a function is continuous at the point in question, substitute the value straight in.
Example: lim(x→ −3) [x² − 13] = (−3)² − 13 = 9 − 13 = −4
Polynomials are continuous everywhere, so direct substitution always works for them.
This identity is foundational and appears constantly in later topics (derivatives of trig functions, Taylor series).
lim(x→0) sin(x)/x = 1
You cannot evaluate this by substitution because it gives 0/0, an indeterminate form.
The result is proved geometrically (squeeze theorem) and simply needs to be memorised.
Scaling the argument:
If the denominator has a constant factor, pull it out.
Example: lim(t→0) sin(t)/(12t) = (1/12) · lim(t→0) sin(t)/t = (1/12)(1) = 1/12
The key move: rewrite the expression so that you have sin(something)/something, then apply the identity.
For a piecewise function, the limit at the boundary depends on what the rules give from each side, not on the function's declared value at that point.
Example: f(x) = { 3x + 7 if x ≠ 2 ; 5 if x = 2 }
From the right: lim(x→2⁺) f(x) = 3(2) + 7 = 13
From the left: lim(x→2⁻) f(x) = 3(2) + 7 = 13
Both sides agree, so lim(x→2) f(x) = 13.
Note: f(2) = 5, which is different from 13. The limit exists but does not equal the function value, so f is not continuous at x = 2.
When the left-hand and right-hand limits give different numbers, the two-sided limit does not exist.
Example: v(t) = { t² − 1 if t > −3 ; 12 + t if t ≤ −3 }
From the right: lim(t→−3⁺) v(t) = (−3)² − 1 = 9 − 1 = 8
From the left: lim(t→−3⁻) v(t) = 12 + (−3) = 9
8 ≠ 9, so lim(t→−3) v(t) does not exist (DNE).
The sides do not agree, and that is the end of the matter. No further work is needed.
\lim_{x \to c} f(x) = L \quad \text{(formal limit notation)}\lim_{x \to 0} \frac{\sin x}{x} = 1\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \quad \text{for any constant } k \neq 0Limit existence rule: lim(x→c) f(x) = L exists if and only if lim(x→c⁻) f(x) = lim(x→c⁺) f(x) = L.
Direct substitution: If f is continuous at c, then lim(x→c) f(x) = f(c).
Students often assume lim(x→c) f(x) must equal f(c). It does not have to. The limit describes approach behaviour, and the function's declared value at c is a separate question entirely. The piecewise example above (limit = 13, f(2) = 5) is a classic exam trap.
Students sometimes plug 0 into sin(x)/x and write "0/0 = undefined." The limit is 1; 0/0 is indeterminate, which means the limit might exist but you need a different technique to find it.
When one side of a piecewise boundary works out, students occasionally stop and report that as the limit. You must always check both sides. If they disagree, the limit is DNE.
Confusing "limit does not exist" with "limit is infinity." A limit that goes to ±∞ also does not exist (in the real-number sense), but the reason is different from the left/right disagreement shown here.
⚠️ Piecewise limit questions are near-guaranteed on exams. The standard move: evaluate the limit from each side, compare, and state whether the limit exists or is DNE. Show both computations.
⚠️ The sin(x)/x = 1 identity comes back repeatedly: in derivative proofs for sin and cos, in L'Hôpital's Rule problems, and in Taylor/Maclaurin series. Memorise it.
⚠️ "Does the limit exist?" and "Is the function continuous?" are different questions. A limit can exist at a point where the function is not continuous (the piecewise example). Exams test whether you can distinguish the two.
⚠️ When asked to evaluate a limit of a piecewise function, always state the left-hand and right-hand limits separately before giving the final answer. Marks are awarded for the process.
True or False: If lim(x→c) f(x) exists, then f(c) must be defined. Answer: False. The limit can exist even if f(c) is undefined.
Fill in the blank: lim(t→0) sin(5t)/(5t) = ___ Answer: 1
True or False: If the left-hand limit equals 7 and the right-hand limit equals 7, the two-sided limit is 7. Answer: True.
Fill in the blank: lim(x→0) sin(x)/(3x) = ___ Answer: 1/3
True or False: lim(x→2) f(x) = f(2) for every function. Answer: False. This holds only when f is continuous at x = 2.
Q: Evaluate lim(x→4) (x² + 2x − 3).
A: Direct substitution. (4)² + 2(4) − 3 = 16 + 8 − 3 = 21.
Q: Evaluate lim(t→0) sin(t)/(7t).
A: Factor out 1/7. (1/7) · lim(t→0) sin(t)/t = (1/7)(1) = 1/7.
Q: Let g(x) = { 2x + 1 if x < 3 ; 10 − x if x ≥ 3 }. Find lim(x→3) g(x), or state that it does not exist.
A: Left: lim(x→3⁻) (2x + 1) = 7. Right: lim(x→3⁺) (10 − x) = 7. Both sides equal 7, so lim(x→3) g(x) = 7.
Q: Let h(x) = { x² if x > 1 ; 3x − 1 if x ≤ 1 }. Does lim(x→1) h(x) exist?
A: Right: lim(x→1⁺) x² = 1. Left: lim(x→1⁻) (3x − 1) = 2. Since 1 ≠ 2, the limit does not exist (DNE).
Q: Suppose lim(x→5) f(x) = 12 but f(5) = 3. Is f continuous at x = 5? Why or why not?
A: No. Continuity requires lim(x→c) f(x) = f(c). Here 12 ≠ 3, so the condition fails.
Limits are the foundation of derivatives. The definition of the derivative is itself a limit: f'(x) = lim(h→0) [f(x+h) − f(x)] / h. Everything in the next recitation (10/03) builds directly on this.
Continuity, which is tested by comparing the limit to the function value, comes up again in the Intermediate Value Theorem and in integration (a function must be reasonably well-behaved for the integral to exist in the standard sense).
The sin(x)/x identity reappears in L'Hôpital's Rule, Taylor series expansions of sin and cos, and Fourier analysis in later courses.
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