# Long Questions

If f(x) = 3x + 2,
a) find f(2.001), f(2.0001), f(1.999), f(1.9999)
b) find $\lim \limits_{x \to 2^-}$f(x), $\lim \limits_{x \to 2^+}$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,
$\lim \limits_{x \to 2^-}$f(x) = $\lim \limits_{x \to 2^-}$3x + 2 = $\lim \limits_{x \to 2^-}$3(2) + 2 = 8

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

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

Functional value at = 3,
f(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) = $\begin{cases} 2 - x^2, &x \leq 2 \\ x - 4, &x > 2 \end{cases}$ at x = 2
b) f(x) = $\begin{cases} x, &x < 0 \\ 0, &x = 0 \\ x^2, &x > 0 \end{cases}$ at x = 0
c) f(x) = $\begin{cases} \frac{2}{5 - x}, &x \leq 3 \\ 5 - x, &x > 3 \end{cases}$ at x = 3
d) f(x) = $\begin{cases} x^2 + 2, &x < 5 \\ 6x - 3, &x = 5 \\ 3x + 12, &x > 5 \end{cases}$ at x = 5

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

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

Right hand limit (RHL) of f(x) at x = 2,
$\lim \limits_{x \to 2^+}$f(x) = $\lim \limits_{x \to 2^+}$ 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) = $\begin{cases} x, &x < 0 \\ 0, &x = 0 \\ x^2, &x > 0 \end{cases}$ at x = 0

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

Right hand limit (RHL) of f(x) at x = 0,
$\lim \limits_{x \to 0^+}$f(x) = $\lim \limits_{x \to 0^+}$ 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) = $\begin{cases} \frac{2}{5 - x}, &x \leq 3 \\ 5 - x, &x > 3 \end{cases}$ at x = 3

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

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

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

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

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

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

Right hand limit (RHL) of f(x) at x = 5,
$\lim \limits_{x \to 3^+}$f(x) = $\lim \limits_{x \to 3^+}$ 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) = $\begin{cases} kx + 3, &x \geq 2 \\ 3x - 1, &x < 2 \end{cases}$
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,
$\lim \limits_{x \to 2^-}$f(x) = $\lim \limits_{x \to 2^-}$3x - 1
= 6 - 1
= 5

Right hand limit (RHL) of f(x) at x = 2,
$\lim \limits_{x \to 2^+}$f(x) = $\lim \limits_{x \to 2^+}$ 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) = $\begin{cases} \frac{2x^2 - 18}{x - 3} &\text{for } x \not = 3 \\ k &\text{for } x = 3 \end{cases}$
Find the value of k so that f(x) is continuous at x = 3.

Here,
Limiting value of f(x) at x = 3,
$\lim \limits_{x \to 3}$f(x) = $\lim \limits_{x \to 3} \frac{2x^2 - 18}{x - 3} (\text{form } \frac 00)$ = $\lim \limits_{x \to 3} \frac{2(x + 3)(x - 3)}{x - 3}$ = 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) = $\begin{cases} 2x - 3, &x < 2 \\ 2, &x = 2 \\ 3x - 5, &x > 2 \end{cases}$
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,
$\lim \limits_{x \to 2^+}$f(x) = $\lim \limits_{x \to 2^+}$ 3x - 5 = 6 - 5
= 1

Left hand limit (LHL) of f(x) at x = 2,
$\lim \limits_{x \to 2^-}$f(x) = $\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) = $\begin{cases} 2x - 3, &x < 2 \\ 1, &x = 2 \\ 3x - 5, &x > 2 \end{cases}$

A function f(x) is defined below:
f(x) = $\begin{cases} 2x^2 + 1, &x < 3 \\ 5, &x = 3 \\ 6x + 1, &x > 3 \end{cases}$
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,
$\lim \limits_{x \to 3^-}$f(x) = $\lim \limits_{x \to 3^-}$2x2 + 1 = 2 × 9 + 1
= 19

Right hand limit (RHL) of f(x) at x = 3,
$\lim \limits_{x \to 3^+}$f(x) = $\lim \limits_{x \to 3^+}$ 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) = $\begin{cases} 2x^2 + 1, &x < 3 \\ 19, &x = 3 \\ 6x + 1, &x > 3 \end{cases}$