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Continuity and Differentiability - Continuity

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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A function f(x)f(x) is said to be continuous at a point x=cx = c if the function is defined at x=cx = c and the limit of the function as xx approaches cc is equal to f(c)f(c).

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Mathematically, ff is continuous at x=cx = c if lim⁑xβ†’cf(x)=f(c)\lim_{x \to c} f(x) = f(c). This implies that the Left Hand Limit (LHL), Right Hand Limit (RHL), and the value of the function at that point must all exist and be equal.

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A function is continuous in an open interval (a,b)(a, b) if it is continuous at every point in that interval.

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A function is continuous in a closed interval [a,b][a, b] if it is continuous in (a,b)(a, b), lim⁑xβ†’a+f(x)=f(a)\lim_{x \to a^+} f(x) = f(a), and lim⁑xβ†’bβˆ’f(x)=f(b)\lim_{x \to b^-} f(x) = f(b).

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If ff and gg are two real functions continuous at a real number cc, then f+gf + g, fβˆ’gf - g, and fβ‹…gf \cdot g are continuous at x=cx = c. The quotient f/gf/g is continuous at x=cx = c provided g(c)β‰ 0g(c) \neq 0.

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Every polynomial function is continuous. Identity functions, constant functions, sine, and cosine functions are continuous everywhere in their domains.

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The composition of two continuous functions is a continuous function. If gg is continuous at cc and ff is continuous at g(c)g(c), then (f∘g)(f \circ g) is continuous at cc.

πŸ“Formulae

lim⁑xβ†’cβˆ’f(x)=lim⁑xβ†’c+f(x)=f(c)\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)

lim⁑hβ†’0f(cβˆ’h)=lim⁑hβ†’0f(c+h)=f(c)\lim_{h \to 0} f(c - h) = \lim_{h \to 0} f(c + h) = f(c)

IfΒ f(x)=\text{If } f(x) = \begin{cases} L(x) & x < c \ R(x) & x \geq c \end{cases},Β thenΒ lim⁑xβ†’cβˆ’L(x)=R(c), \text{ then } \lim_{x \to c^-} L(x) = R(c)

πŸ’‘Examples

Problem 1:

Check the continuity of the function ff given by f(x)=2x+3f(x) = 2x + 3 at x=1x = 1.

Solution:

  1. Find the value of the function at x=1x = 1: f(1)=2(1)+3=5f(1) = 2(1) + 3 = 5.
  2. Find the limit as xβ†’1x \to 1: lim⁑xβ†’1(2x+3)=2(1)+3=5\lim_{x \to 1} (2x + 3) = 2(1) + 3 = 5.
  3. Since lim⁑xβ†’1f(x)=f(1)\lim_{x \to 1} f(x) = f(1), the function is continuous at x=1x = 1.

Explanation:

According to the definition of continuity, if the limit of the function as it approaches a point equals the function's value at that point, the function is continuous there.

Problem 2:

Find the value of kk so that the function f(x)={kx2ifΒ x≀23ifΒ x>2f(x) = \begin{cases} kx^2 & \text{if } x \leq 2 \\ 3 & \text{if } x > 2 \end{cases} is continuous at x=2x = 2.

Solution:

For f(x)f(x) to be continuous at x=2x = 2, we must have LHL=RHL=f(2)LHL = RHL = f(2).

  1. f(2)=k(2)2=4kf(2) = k(2)^2 = 4k.
  2. LHL=lim⁑xβ†’2βˆ’kx2=4kLHL = \lim_{x \to 2^-} kx^2 = 4k.
  3. RHL=lim⁑xβ†’2+3=3RHL = \lim_{x \to 2^+} 3 = 3.
  4. Set LHL=RHLLHL = RHL: 4k=3β‡’k=344k = 3 \Rightarrow k = \frac{3}{4}.

Explanation:

In piecewise functions, we equate the limits from both sides of the point of interest to the value of the function at that point to ensure there is no jump or break in the graph.