Continuity of a Function

1. Statement of the Concept — What is Continuity of a Function?

A function [f(x)] is continuous on an interval if it is continuous at every point of that interval.
Formally, [f(x)] is continuous at [x = a] if:

1. [f(a)] is defined.
2. [\displaystyle \lim_{x \to a} f(x)] exists.
3. [\displaystyle \lim_{x \to a} f(x) = f(a)].

If these three hold for every [a] in an interval, [f] is continuous on that interval.

(For more on limits used here, see Introduction to Continuity.)

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2. Clear Explanation and Reasoning

• Intuition: Continuity at [x = a] means the value the function approaches when [x] gets arbitrarily close to [a] equals the actual value at [a]. There is no “break” at [a].
• Left & right limits: Saying [\displaystyle \lim_{x \to a} f(x)] exists means both left-hand limit [\lim_{x \to a^-} f(x)] and right-hand limit [\lim_{x \to a^+} f(x)] exist and are equal.
• Types of continuity on intervals:
  • Continuous on [(a, b)] (closed interval): continuous at every point of [a, b], with special endpoint definitions: right-hand limit at [a] equals [f(a)], left-hand limit at [b] equals [f(b)].
  • Continuous on ]a, b[ (open interval): continuous at every interior point.
• Important consequence: If two functions [f] and [g] are continuous at [a], then so are [f+g], [f-g], [f\cdot g], and — if [g(a) \ne 0] — [\dfrac{f}{g}]. Also compositions [f(g(x))] are continuous where defined.

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3. Key Features (Quick Facts)

• Polynomials are continuous for all real numbers.
• Rational functions are continuous wherever the denominator ≠ 0.
• Trigonometric, exponential, and logarithmic functions are continuous on their domains.
• Continuity is a local property — it’s checked pointwise.
• A function can be continuous everywhere but nowhere differentiable (e.g., Weierstrass function — advanced topic).
• Differentiability ⇒ continuity, but continuity ⇏ differentiability.

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4. Important Formulas / Rules (Table)

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5. Conceptual Questions with solution

1. If [f(x)] is continuous at [x = 2] and [f(2)=5], what is [\displaystyle \lim_{x \to 2} f(x)]?

**Solution:** By definition of continuity at [2], the limit equals the function value: [\displaystyle \lim_{x \to 2} f(x)=f(2)=5].

2. Is [f(x)=\dfrac{x^2-1}{x-1}] continuous at [x=1]?

**Solution:** Simplify for [x\ne 1]: [\dfrac{x^2-1}{x-1}=\dfrac{(x-1)(x+1)}{x-1}=x+1]. So for [x\ne 1], [f(x)=x+1]. The limit as [x\to1] is [2]. But [f(1)] is undefined (original expression has denominator 0). So continuity fails at [1]. Therefore **not continuous at [x=1]**.

3. If [f] and [g] are continuous at [a], is [h(x)=5f(x)-3g(x)] continuous at [a]?

**Solution:** Yes. Linear combinations of continuous functions are continuous. So [h] is continuous at [a].

4. Is [f(x)=\sin x] continuous for all real numbers?

**Solution:** Yes. Sine is continuous for all real numbers (trigonometric functions are continuous on their domains, and domain of sine is [\mathbb{R}]).

5. Can a function be continuous on (0,1) but not continuous on [0,1]?

**Solution:** Yes — for example [f(x)=\dfrac{1}{x}] is continuous on (0,1) but not at 0, so not continuous on [0,1]. Endpoints matter.

6. Is [f(x)=|x|] continuous at [x=0]?

**Solution:** Check limit: [\lim_{x\to0^-}|x|=0], [\lim_{x\to0^+}|x|=0], and [f(0)=0]. All equal ⇒ continuous at 0. (But not differentiable at 0.)

7. If [\lim_{x\to a} f(x)=L] and [f(a)=M] with [L\ne M], is [f] continuous at [a]?

**Solution:** No. Continuity requires [L=f(a)]; mismatch ⇒ discontinuous.

8. Are exponential functions [e^x] continuous for all real x?

**Solution:** Yes. [e^x] is continuous on [\mathbb{R}].

9. If [f(x)=\begin{cases}x^2, & x\le 1\\ 2x+1, & x>1\end{cases}], is [f] continuous at [x=1]?

**Solution:** Compute left limit: [\lim_{x\to1^-}x^2=1]. Right limit: [\lim_{x\to1^+}2x+1=3]. They are different → limit does not exist → not continuous at 1.

10. If [f] continuous at [a] and [g] continuous at [f(a)], is [g(f(x))] continuous at [a]?

**Solution:** Yes. Composition of continuous functions is continuous where defined.

11. A function has [\lim_{x\to 0} f(x)=3] but [f(0)=3]. Is the function continuous at 0?

**Solution:** Yes — the limit equals the function value at 0, so it’s continuous at 0.

12. If [f(x) = \dfrac{\sin x}{x}] with [f(0)] defined as [1], is [f] continuous at 0?

**Solution:** We know [\lim_{x\to0} \dfrac{\sin x}{x} = 1]. If we define [f(0)=1], then the limit equals the value → continuous at 0. (This is a removable discontinuity resolved by defining the function value appropriately.)

13. Can a function be continuous everywhere but not continuous at infinity?

**Solution:** “Continuity at infinity” is a different concept. A function can be continuous on all real numbers but may not have a limit as [x\to\infty]. That does not contradict continuity on real line.

14. Is [f(x)=\begin{cases} \sin(\dfrac{1}{x}), & x\ne 0\\ 0, & x=0\end{cases}] continuous at 0?

**Solution:** [\sin(\dfrac{1}{x})] oscillates between [-1,1] as [x\to0], so the limit does not exist. But [f(0)=0]. Since limit ≠ value (limit doesn’t exist), function is **not continuous** at 0.

15. If [f] continuous on [a, b] and [f(a)\ne f(b)], does [f] take every value between [f(a)] and [f(b)]?

**Solution:** Yes — by the Intermediate Value Theorem (IVT). If [f] is continuous on [a,b], it attains every value between [f(a)] and [f(b)]. (Important theorem — used in root-finding and existence proofs.)

16. If [f(x)] is continuous and non-zero at [a], can it be zero at points arbitrarily close to [a]?

**Solution:** If [f(a)\ne 0], by continuity there’s a small neighborhood around [a] where [f(x)] keeps the same sign and doesn’t become 0. So it cannot be zero arbitrarily close if it’s non-zero at [a].

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6. FAQ / Common Misconceptions

1. “If limit exists, function is continuous.” — True or False?

**Answer:** False. Continuity also requires the function to be defined at that point and the function value to equal the limit. Example: [\dfrac{x-1}{x-1}] has limit 1 at [x=1] but is undefined at [1].

2. “Continuity at endpoints of [a,b] requires both left and right limits.” — Clarify.

**Answer:** At left endpoint [a] only right-hand limit matters; at right endpoint [b] only left-hand limit matters. You do not require both-sided limits at endpoints of closed intervals.

3. “Differentiable ⇒ continuous, so continuity implies differentiability.” — True?

**Answer:** No. The correct implication is differentiable ⇒ continuous. The converse is false. Example: [f(x)=|x|] is continuous at 0 but not differentiable there.

4. “A function with a hole is always non-continuous.” — Can it be fixed?

**Answer:** A removable discontinuity (a hole) can be fixed by redefining the function value at that point to equal the limit. After redefining, the function becomes continuous there.

5. “Jump discontinuity vs removable discontinuity — difference?”

**Answer:** A removable discontinuity has both one-sided limits equal but function value different or undefined. A jump has left and right limits unequal.

6. “If both one-sided limits exist, function is continuous.” — Is that enough?

**Answer:** Not enough. They must be equal and equal to the function value at that point.

7. “Rational functions are always continuous.” — True?

**Answer:** False. Rational functions are continuous on their domain (where denominator ≠ 0), but not at points where denominator is zero.

8. “If limit is infinite, function is continuous.” — Explain.

**Answer:** No. If the limit is infinite (vertical asymptote), the limit does not exist as a finite number, so continuity fails at that point.

9. “Continuity means graph is smooth.” — Is smooth same as continuous?

**Answer:** Not exactly. “Smooth” usually means differentiable (and sometimes many times differentiable). Continuous means no breaks; it can have corners (like |x|) yet be continuous.

10. “If function equals different formulas on two sides and they match numerically at the point, it is continuous.” — Any caveats?

**Answer:** If both one-sided limits equal the same number and that number equals the function value, then yes it’s continuous. But you must check function value is defined there.

11. “A function continuous everywhere must have a finite limit at infinity.” — True?

**Answer:** False. Example: [\sin x] is continuous everywhere but does not have a limit as [x\to\infty].

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8. Practice Questions With Step-by-Step Solutions

Practice 1. Check continuity of [f(x)=x^3-2x+1] at [x=1].
Solution (step-by-step):

1. This is a polynomial → continuous everywhere.
2. Therefore continuous at [x=1].
   (Alternative computation: [f(1)=1-2+1=0], [\lim_{x\to1}x^3-2x+1=0] → equal.)

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Practice 2. Check continuity of [f(x)=\dfrac{x^2-4}{x-2}] at [x=2].
Solution:

1. For [x\ne2], simplify: [\dfrac{(x-2)(x+2)}{x-2}=x+2].
2. Limit as [x\to2] is [4].
3. But original function is undefined at [x=2] (denominator zero).
4. Hence discontinuous at [2]. (Removable discontinuity; define [f(2)=4] to make continuous.)

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Practice 3. Determine continuity at 0 of [f(x)=\begin{cases}\dfrac{\sin x}{x}, & x\ne 0\ 1, & x=0\end{cases}].
Solution:

1. Known limit: [\lim_{x\to0}\dfrac{\sin x}{x}=1].
2. [f(0)=1].
3. Limit equals value → continuous at 0.

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Practice 4. Check continuity at [x=0] of [f(x)=\begin{cases}x\sin(\dfrac{1}{x}), & x\ne0\ 0, & x=0\end{cases}].
Solution:

1. For [x\ne0], [|x\sin(\dfrac{1}{x})|\le|x|\cdot 1=|x|].
2. So [\lim_{x\to0}x\sin(\dfrac{1}{x})=0] by squeeze theorem.
3. [f(0)=0]. Hence continuous at 0.

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Practice 5. Is [ \displaystyle f(x)=\left\{ \begin{array}{l}\begin{array}{*{20}{l}} {\dfrac{{{{x}^{2}}-9}}{{x-3}},} & {x\ne 3} & {} \end{array}\\6,\ \ x=3\ \end{array} \right.] continuous at 3?
Solution:

1. Simplify for [x\ne3]: [\dfrac{(x-3)(x+3)}{x-3}=x+3].
2. Limit as [x\to3] is [6].
3. [f(3)=6]. So continuous at 3.

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Practice 6. Determine continuity of [f(x)=\dfrac{|x|}{x}] at [x=0] and elsewhere.
Solution:

1. For [x>0], [\dfrac{|x|}{x}=\dfrac{x}{x}=1].
2. For [x<0], equals [-1].
3. So left-hand limit at 0 is [-1], right-hand limit is [1]. They differ → limit does not exist → discontinuous at 0.
4. For [x\ne0], it’s constant on each side → continuous everywhere except 0.

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Practice 7. Check continuity of [f(x)=\ln x] at [x=1] and at [x\le0].
Solution:

1. Domain of [\ln x] is (0, ∞). So not defined for [x\le0].
2. At [x=1], [\ln 1=0] and logarithm is continuous on its domain → continuous at 1.
3. For points ≤ 0, function not defined → not continuous there.

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Practice 8. Let [f(x)=\begin{cases}x^2\sin(\dfrac{1}{x}), & x\ne 0\ 0, & x=0\end{cases}]. Show continuity at 0.
Solution:

1. [|x^2\sin(\dfrac{1}{x})|\le x^2].
2. [\lim_{x\to0}x^2=0] so by squeeze theorem limit is 0.
3. [f(0)=0] → continuous.

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Practice 9. Is [f(x)=\begin{cases}\dfrac{x^2-1}{x-1}, & x\ne1\ 3, & x=1\end{cases}] continuous at 1?
Solution:

1. Simplify for [x\ne1]: becomes [x+1]. Limit as [x\to1] is [2].
2. [f(1)=3]. Since limit ≠ function value, discontinuous at 1.

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Practice 10. Determine continuity of [f(x)=\begin{cases}\sin x, & x\le \pi\ 2, & x> \pi\end{cases}] at [x=\pi].
Solution:

1. Left-hand limit as [x\to\pi^-] is [\sin\pi=0].
2. Right-hand limit as [x\to\pi^+] is [2]. They differ → limit does not exist.
3. So discontinuous at [x=\pi].

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Quick Revision Box (Exam Tips)

• To check continuity at [a]:
  1. See if [f(a)] is defined.
  2. Compute left & right limits (or overall limit).
  3. Check if limit = [f(a)].
• For piecewise functions, always compute one-sided limits.
• Recognize standard continuous families: polynomials, trig (on domain), exponentials, logs (on domain), rational (on domain).
• Removable discontinuities can be fixed by redefining value at that point.