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Long Questions

If f(x) = 3x + 2,
a) find f(2.001), f(2.0001), f(1.999), f(1.9999)
b) find lim⁡x→2−\lim \limits_{x \to 2^-}f(x), lim⁡x→2+\lim \limits_{x \to 2^+}f(x) and f(2).
c) Is it continuous at x = 2?

Here,
f(x) = 3x + 2
  1. f(2.001) = 3(2.001) + 2 = 8.003
    f(2.0001) = 3(2.0001) + 2 = 8.0003
    f(1.999) = 3(1.999) + 2 = 7.997
    f(1.9999) = 3(1.9999) + 2 = 7.9997

  2. Left hand limit at x = 2,
    lim⁡x→2−\lim \limits_{x \to 2^-}f(x) = lim⁡x→2−\lim \limits_{x \to 2^-}3x + 2 = lim⁡x→2−\lim \limits_{x \to 2^-}3(2) + 2 = 8

    Right hand limit at x = 2,
    lim⁡x→2+\lim \limits_{x \to 2^+}f(x) = lim⁡x→2+\lim \limits_{x \to 2^+}3x + 2 = lim⁡x→2+\lim \limits_{x \to 2^+}3(2) + 2 = 8

    Functional value at = 2,
    f(2) = 3(2) + 2 = 8

  3. Left hand limit = Right hand limit = f(2). So, yes it is continuous at x = 2.

If f(x) = x2−9x−3\frac{x^2 - 9}{x - 3}, find:
a) f(2.999) and f(3.001)
b) f(2.999) and f(3.001) equal after approximation
c) Discuss about continuity at x = 3

Here,
f(x) = x2−9x−3\frac{x^2 - 9}{x - 3}
  1. f(2.999) = (2.999)2−92.999−3\frac{(2.999)^2 - 9}{2.999 - 3} = 5.999
    f(3.001) = (3.001)2−93.001−3\frac{(3.001)^2 - 9}{3.001 - 3} = 6.001
  2. f(2.999) = (2.999)2−92.999−3\frac{(2.999)^2 - 9}{2.999 - 3} = 5.999 ≈ 6
    f(3.001) = (3.001)2−93.001−3\frac{(3.001)^2 - 9}{3.001 - 3} = 6.001 ≈ 6
    So, f(3.001) equals to f(3.001) after approximation.
  3. Left hand limit at x = 3,
    lim⁡x→3−\lim \limits_{x \to 3^-}f(x) = lim⁡x→3−x2−9x−3(form 00)\lim \limits_{x \to 3^-} \frac{x^2 - 9}{x - 3} ( \text{form } \frac 00 ) = lim⁡x→3−(x+3)(x−3)x−3\lim \limits_{x \to 3^-} \frac{(x + 3)(x - 3)}{x - 3}
    = lim⁡x→3−\lim \limits_{x \to 3^-}(x + 3)
    = 6

    Right hand limit at x = 3,
    lim⁡x→3+\lim \limits_{x \to 3^+}f(x) = lim⁡x→3+x2−9x−3(form 00)\lim \limits_{x \to 3^+} \frac{x^2 - 9}{x - 3} ( \text{form } \frac 00 ) = lim⁡x→3+(x+3)(x−3)x−3\lim \limits_{x \to 3^+} \frac{(x + 3)(x - 3)}{x - 3}
    = lim⁡x→3+\lim \limits_{x \to 3^+}(x + 3)
    = 6

    Functional value at = 3,
    f(3) = (x+3)(x−3)x−3\frac{(x + 3)(x - 3)}{x - 3} = 3 + 3 = 6

    Left hand limit = Right hand limit = f(3). So, it is continuous at x = 3.

Examine the continuity or discontinuity at points mentioned below:
a) f(x) = {2−x2,x≤2x−4,x>2\begin{cases} 2 - x^2, &x \leq 2 \\ x - 4, &x > 2 \end{cases} at x = 2
b) f(x) = {x,x<00,x=0x2,x>0\begin{cases} x, &x < 0 \\ 0, &x = 0 \\ x^2, &x > 0 \end{cases} at x = 0
c) f(x) = {25−x,x≤35−x,x>3\begin{cases} \frac{2}{5 - x}, &x \leq 3 \\ 5 - x, &x > 3 \end{cases} at x = 3
d) f(x) = {x2+2,x<56x−3,x=53x+12,x>5\begin{cases} x^2 + 2, &x < 5 \\ 6x - 3, &x = 5 \\ 3x + 12, &x > 5 \end{cases} at x = 5

  1. f(x) = {2−x2,x≤2x−4,x>2\begin{cases} 2 - x^2, &x \leq 2 \\ x - 4, &x > 2 \end{cases} at x = 2

    Here,
    Left hand limit (LHL) of f(x) at x = 2,
    lim⁡x→2−\lim \limits_{x \to 2^-}f(x) = lim⁡x→2−\lim \limits_{x \to 2^-} 2 - x2 = 2 - 4
    = -2

    Right hand limit (RHL) of f(x) at x = 2,
    lim⁡x→2+\lim \limits_{x \to 2^+}f(x) = lim⁡x→2+\lim \limits_{x \to 2^+} x - 4 = 2 - 4
    = -2

    Functional value at x = 2,
    f(x) = 2 - x2
    so, f(2) = 2 - 4 = -2

    From above, LHL = RHL = f(2) = -2. So, f(x) is continuous at x = 2.

  2. f(x) = {x,x<00,x=0x2,x>0\begin{cases} x, &x < 0 \\ 0, &x = 0 \\ x^2, &x > 0 \end{cases} at x = 0

    Here,
    Left hand limit (LHL) of f(x) at x = 0,
    lim⁡x→0−\lim \limits_{x \to 0^-}f(x) = lim⁡x→0−\lim \limits_{x \to 0^-} x = 0

    Right hand limit (RHL) of f(x) at x = 0,
    lim⁡x→0+\lim \limits_{x \to 0^+}f(x) = lim⁡x→0+\lim \limits_{x \to 0^+} x2 = 0

    Functional value at x = 0,
    f(x) = 0
    so, f(0) = 0

    From above, LHL = RHL = f(0) = 0. So, f(x) is continuous at x = 0.

  3. f(x) = {25−x,x≤35−x,x>3\begin{cases} \frac{2}{5 - x}, &x \leq 3 \\ 5 - x, &x > 3 \end{cases} at x = 3

    Here,
    Left hand limit (LHL) of f(x) at x = 3,
    lim⁡x→3−\lim \limits_{x \to 3^-}f(x) = lim⁡x→3−25−x\lim \limits_{x \to 3^-} \frac{2}{5 - x} = 25−3\frac{2}{5 - 3}
    = 1

    Right hand limit (RHL) of f(x) at x = 3,
    lim⁡x→3+\lim \limits_{x \to 3^+}f(x) = lim⁡x→3+\lim \limits_{x \to 3^+} 5 - x
    = 5 - 3
    = 2

    Functional value at x = 3,
    f(x) = 25−x\frac{2}{5 - x}
    so, f(3) = 25−3\frac{2}{5 - 3} = 1

    From above, RHL ≠ LHL = f(3). So, f(x) is discontinuous at x = 3.

  4. f(x) = {x2+2,x<56x−3,x=53x+12,x>5\begin{cases} x^2 + 2, &x < 5 \\ 6x - 3, &x = 5 \\ 3x + 12, &x > 5 \end{cases} at x = 5

    Here,
    Left hand limit (LHL) of f(x) at x = 5,
    lim⁡x→3−\lim \limits_{x \to 3^-}f(x) = lim⁡x→3−\lim \limits_{x \to 3^-} x2 + 2 = 52 + 2
    = 27

    Right hand limit (RHL) of f(x) at x = 5,
    lim⁡x→3+\lim \limits_{x \to 3^+}f(x) = lim⁡x→3+\lim \limits_{x \to 3^+} 3x + 12 = 15 + 12
    = 27

    Functional value at x = 5,
    f(x) = 6x - 3
    so, f(3) = 30 - 3 = 27

    From above, LHL = RHL = f(5) = 27. So, f(x) is discontinuous at x = 5.

A function f(x) is defined below:
f(x) = {kx+3,x≥23x−1,x<2\begin{cases} kx + 3, &x \geq 2 \\ 3x - 1, &x < 2 \end{cases}
Find the value of k so that f(x) is continuous at x = 2.

Here,
Left hand limit (LHL) of f(x) at x = 2,
lim⁡x→2−\lim \limits_{x \to 2^-}f(x) = lim⁡x→2−\lim \limits_{x \to 2^-} 3x - 1
= 6 - 1
= 5

Right hand limit (RHL) of f(x) at x = 2,
lim⁡x→2+\lim \limits_{x \to 2^+}f(x) = lim⁡x→2+\lim \limits_{x \to 2^+} kx + 3 = 2k + 3

Functional value at x = 2,
f(x) = kx + 3
so, f(2) = 2k + 3

Since, f(x) is continuous, LHL = RHL = f(2) or, 5 = 2k + 3 = 2k + 3

Taking, 5 = 2k + 3 or, 2 = 2k
∴ k = 1.

A function f(x) is defined below:
f(x) = {2x2−18x−3for x≠3kfor x=3\begin{cases} \frac{2x^2 - 18}{x - 3} &\text{for } x \not = 3 \\ k &\text{for } x = 3 \end{cases}
Find the value of k so that f(x) is continuous at x = 3.

Here,
Limiting value of f(x) at x = 3,
lim⁡x→3\lim \limits_{x \to 3}f(x) = lim⁡x→32x2−18x−3(form 00)\lim \limits_{x \to 3} \frac{2x^2 - 18}{x - 3} (\text{form } \frac 00) = lim⁡x→32(x+3)(x−3)x−3\lim \limits_{x \to 3} \frac{2(x + 3)(x - 3)}{x - 3} = 2(3 + 3) = 12

Functional value at x = 3, f(x) = k so, f(3) = k

Since, f(x) is continuous, Limiting value = f(3) or, 12 = 5
∴ k = 12.

A function f(x) is defined below:
f(x) = {2x−3,x<22,x=23x−5,x>2\begin{cases} 2x - 3, &x < 2 \\ 2, &x = 2 \\ 3x - 5, &x > 2 \end{cases}
Is the function f(x) continuous at x = 2? If not, how can the function f(x) be made continuous at x = 2?

Here,
Right hand limit (RHL) of f(x) at x = 2,
lim⁡x→2+\lim \limits_{x \to 2^+}f(x) = lim⁡x→2+\lim \limits_{x \to 2^+} 3x - 5 = 6 - 5
= 1

Left hand limit (LHL) of f(x) at x = 2,
lim⁡x→2−\lim \limits_{x \to 2^-}f(x) = lim⁡x→2−\lim \limits_{x \to 2^-} 2x - 3 = 4 - 3
= 1

Functional value at x = 2,
f(x) = f(2) = 2
LHL = RHL ≠ f(2). So, f(x) is not continuous.

To make f(x) continuous, we redefined the function as follows:
f(x) = {2x−3,x<21,x=23x−5,x>2\begin{cases} 2x - 3, &x < 2 \\ 1, &x = 2 \\ 3x - 5, &x > 2 \end{cases}

A function f(x) is defined below:
f(x) = {2x2+1,x<35,x=36x+1,x>3\begin{cases} 2x^2 + 1, &x < 3 \\ 5, &x = 3 \\ 6x + 1, &x > 3 \end{cases}
Is the function f(x) continuous at x = 3? If not, how can the function f(x) be made continuous?

Here,

Left hand limit (LHL) of f(x) at x = 3,
lim⁡x→3−\lim \limits_{x \to 3^-}f(x) = lim⁡x→3−\lim \limits_{x \to 3^-} 2x2 + 1 = 2 × 9 + 1
= 19

Right hand limit (RHL) of f(x) at x = 3,
lim⁡x→3+\lim \limits_{x \to 3^+}f(x) = lim⁡x→3+\lim \limits_{x \to 3^+} 6x + 1 = 6 × 3 + 1
= 19

Functional value at x = 3,
f(x) = f(3) = 5
LHL = RHL ≠ f(2). So, f(x) is not continuous.

To make f(x) continuous, we redefined the function as follows:
f(x) = {2x2+1,x<319,x=36x+1,x>3\begin{cases} 2x^2 + 1, &x < 3 \\ 19, &x = 3 \\ 6x + 1, &x > 3 \end{cases}

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