Higher Secondary Mathematics: 21.2 - Solved Examples on Continuity

In the previous section, we saw continuous functions. We also saw some solved examples and the concept of infinity. In this section, we will see a few more solved examples.

Solved example 21.10
Discuss the continuity of the function f defined by
$f(x) = \begin{cases} x+2, & \text{if}~x\le 1 \\[1.5ex] x-2, & \text{if}~x>1 \end{cases}$
Solution:
• The given function has two segments.
(i) When input x is less than or equal to 1, we must use the segment f(x) = x+2
(ii) When input x is greater than 1, we must use the segment f(x) = x−2
• So x = 1 is a critical point. We must use this point for analysis. We must consider three cases:
   ♦ Case 1, where input x is less than 1
   ♦ Case 2, where input x is greater than 1
   ♦ Case 3, where input x is equal to 1
These three cases will cover all real numbers.

Case 1: x <1
1. Consider any arbitrary point c such that c < 1. Let us check whether the function is continuous at c.
Since c < 1, we must use the segment f(x) = x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c+2
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point less than 1. So it can be any real number less than 1.
• Therefore we can write:
The given function is continuous at every real number less than 1.

Case 2: x > 1
1. Consider any arbitrary point c such that c > 1. Let us check whether the function is continuous at c.
Since c > 1, we must use the segment f(x) = x−2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c−2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c−2
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point greater than 1. So it can be any real number greater than 1.
• Therefore we can write:
The given function is continuous at every real number greater than 1.

Case 3: x = 1
1. Here we cannot consider an arbitrary point. We must check the continuity at the exact point where x = 1.
2. Applying condition (i):
Since x = 1 is a critical point, we must check both left side limit and right side limit.
◼ Checking left side limit:
• While checking left side limit, all input x values are less than 1. So we must use the segment f(x) = x+2
• Therefore, $\lim_{x\rightarrow 1^{−}} f(x) = (1+2) = 3$
◼ Checking right side limit:
• While checking right side limit, all input x values are greater than 1. So we must use the segment f(x) = x-2
• Therefore, $\lim_{x\rightarrow 1^{+}} f(x) = (1-2) = -1$
◼ The two limits are not the same. So we can write:
$\lim_{x\rightarrow 1} f(x)$ does not exist.
• So this function does not satisfy the first condition at x =1.
3. We can establish continuity only if both conditions are satisfied. Therefore, this function is not continuous at x = 1.

Conclusion:
From cases 1 and 2, we established continuity at all points other than 1.
• So we can write:
For the given function, x = 1 is the only point of discontinuity.

The above steps will become more clear if we use a graph. It is shown in fig.21.6 below:

Fig.21.6

1. For case 1, we use the left segment. We see that, for any point where x<1, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>1, the function is continuous.
3. For case 3, we use the two points (1,3) and (1,-1).
   ♦ Left side limit is indicated by (1,3).
   ♦ Right side limit is indicated by (1,-1).
4. We see that, the graph cannot be drawn in a single stroke. At (1,3), we have to lift the pen from the plane of the paper. Then continue drawing from (1,-1).
5. Note the type of circles used for marking the two points.
   ♦ (1,3) is marked with a filled circle.
   ♦ (1,-1) is marked with an ordinary circle.

Solved example 21.11
Find all the points of discontinuity of the function f defined by
$f(x) = \begin{cases} x+2, & \text{if}~x< 1
\\[1.5ex] 0, & \text{if}~x=1
\\[1.5ex] x-2, & \text{if}~x>1
\end{cases}$
Solution:
• The given function has three segments.
(i) When input x is less than 1, we must use the segment f(x) = x+2
(ii) When input x is equal to 1, we must use the segment f(x) = 0
(iii) When input x is greater than 1, we must use the segment f(x) = x−2
• So x = 1 is a critical point. We must use this point for analysis. We must consider three cases:
   ♦ Case 1, where input x is less than 1
   ♦ Case 2, where input x is greater than 1
   ♦ Case 3, where input x is equal to 1
These three cases will cover all real numbers.

Conclusion:
From cases 1 and 2, we established continuity at all points other than 1.
• So we can write:
For the given function, x = 1 is the only point of discontinuity.

• It may be noted that, in the previous example,
$\lim_{x\rightarrow 1^{−}} f(x) ~\ne~ \lim_{x\rightarrow 1^{+}} f(x)$
But $\lim_{x\rightarrow 1^{−}} f(x) ~=~ f(1)$
• In the present example:
$\lim_{x\rightarrow 1^{−}} f(x) ~\ne~ \lim_{x\rightarrow 1^{+}} f(x) ~\ne~ f(1)$.
This is because, for the present example, it is given that: f(1) = 0.

The above steps will become more clear if we use a graph. It is shown in fig.21.7 below:

Fig.21.7

1. For case 1, we use the left segment. We see that, for any point where x<1, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>1, the function is continuous.
3. For case 3, we use the two points (1,3) and (1,-1).
   ♦ Left side limit is indicated by (1,3).
   ♦ Right side limit is indicated by (1,-1).
4. We see that, the graph cannot be drawn in a single stroke. At (1,3), we have to lift the pen from the plane of the paper. Then continue drawing from (1,-1).
5. Note the type of circles used for marking the three points.
   ♦ (1,3) is marked with a ordinary circle.
   ♦ (1,-1) is marked with an ordinary circle.
   ♦ (1,0) is marked with a filled circle.

Solved example 21.12
Discuss the continuity of the function f defined by
$f(x) = \begin{cases} x+2, & \text{if}~x < 0 \\[1.5ex] -x+2, & \text{if}~x>0 \end{cases}$
Solution:
• The given function has two segments.
(i) When input x is less than 0, we must use the segment f(x) = x+2
(ii) When input x is greater than 0, we must use the segment f(x) = −x+2
• So x = 0 is a critical point. We must use this point for analysis. We must consider two cases:
♦ Case 1, where input x is less than zero.
♦ Case 2, where input x is greater than zero.
These two cases will cover all real numbers except zero.

Case 1: x < 0
1. Consider any arbitrary point c such that c < 0. Let us check whether the function is continuous at c.
Since c < 0, we must use the segment f(x) = x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c+2
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point less than 0. So it can be any real number less than 0.
• Therefore we can write:
The given function is continuous at every real number less than 0.

Case 2: x > 0
1. Consider any arbitrary point c such that c > 0. Let us check whether the function is continuous at c.
Since c > 0, we must use the segment f(x) = −x+2
2. Applying condition (i):
• Limiting value of f(x) at x = c is: −c+2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = −c+2
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point greater than 0. So it can be any real number greater than 0.
• Therefore we can write:
The given function is continuous at every real number greater than 0.

Conclusion:
From cases 1 and 2, we established continuity at all points other than zero.
• We need not consider the continuity at zero because, it is not defined in the given function.
• Domain of the given function is: R − {0}
• So we can write:
Since f is continuous at all points in the domain, it is a continuous function.

The above steps will become more clear if we use a graph. It is shown in fig.21.8 below:

Fig.21.8

1. For case 1, we use the left segment. We see that, for any point where x<0, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>0, the function is continuous.
3. We need not consider the case when x = 0.
4. Note the type of circles used for marking the point where x = 0. It is an ordinary circle. The ordinary circle indicates that, the point is not included in the graph.
5. We see that, the graph cannot be drawn in a single stroke. At x=0, we have to lift the pen from the plane of the paper. But x=0 is not defined. Zero is not present in the domain.
• So we can write:
The given function f is continuous for all points in the domain of f.

Solved example 21.13
Discuss the continuity of the function f defined by
$f(x) = \begin{cases} x, & \text{if}~x \ge 0 \\[1.5ex] x^2, & \text{if}~x < 0 \end{cases}$
Solution:
• The given function has two segments.
(i) When input x is less than zero, we must use the segment f(x) = x².
(ii) When input x is greater than or equal to zero, we must use the segment f(x) = x.
• So x = 0 is a critical point. We must use this point for analysis. We must consider three cases:
   ♦ Case 1, where input x is less than 0
   ♦ Case 2, where input x is greater than 0
   ♦ Case 3, where input x is equal to 0
These three cases will cover all real numbers.

Case 1: x < 0
1. Consider any arbitrary point c such that c < 0. Let us check whether the function is continuous at c.
Since c < 0, we must use the segment f(x) = x².
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c².
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c².
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point less than 0. So it can be any real number less than 0.
• Therefore we can write:
The given function is continuous at every real number less than zero.

Case 2: x > 0
1. Consider any arbitrary point c such that c > 0. Let us check whether the function is continuous at c.
Since c > 0, we must use the segment f(x) = x
2. Applying condition (i):
• Limiting value of f(x) at x = c is: c
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = c
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point greater than 0. So it can be any real number greater than 0.
• Therefore we can write:
The given function is continuous at every real number greater than zero.

Case 3: x = 0
1. Here we cannot consider an arbitrary point. We must check the continuity at the exact point where x = 0.
2. Applying condition (i):
Since x = 0 is a critical point, we must check both left side limit and right side limit.
◼ Checking left side limit:
• While checking left side limit, all input x values are less than 0. So we must use the segment f(x) = x².
• Therefore, $\lim_{x\rightarrow 1^{−}} f(x) = (0^2) = 0$
◼ Checking right side limit:
• While checking right side limit, all input x values are greater than 0. So we must use the segment f(x) = x
• Therefore, $\lim_{x\rightarrow 1^{+}} f(x) = 0$
◼ The two limits are the same. So we can write:
$\lim_{x\rightarrow 0} f(x)$ exists.
• So this function satisfies the first condition at x = 0.
3. Applying condition (ii):
When x = 0, we must use the segment f(x) = x.
So we get: f(0) = 0
• We see that: $\lim_{x\rightarrow 0} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = 0.

Conclusion:
From cases 1, 2 and 3, we established continuity at all real numbers.
• The domain of the function is R. So we can write:
The given function f, is continuous at all points in the domain of f. So it is a continuous function.

The above steps will become more clear if we use a graph. It is shown in fig.21.9 below:

Fig.21.9

1. For case 1, we use the left segment. We see that, for any point where x<0, the function is continuous.
2. For case 2, we use the right segment. We see that, for any point where x>0, the function is continuous.
3. For case 3, we use the point (0,0).
♦ Left side limit is zero.
♦ Right side limit is zero.
4. We see that, the graph can be drawn in a single stroke. We do not have to lift the pen from the plane of the paper.

Solved example 21.13
Show that every polynomial function is continuous.
Solution:
1. A function p is polynomial function if it is defined by:
p(x) = a₀ + a₁x + a₂x² + a₃x³ + . . . + a_nxⁿ.
   ♦ a, a₁, a₂ etc., must be real numbers.
   ♦ n must be a natural number.
(Details can be seen here)
2. Many functions that we analyzed in this chapter are polynomial functions.
• f(x) = x is a polynomial function.
   ♦ Here a₀ = 0 and a₁ = 1
• f(x) = x² is a polynomial function.
   ♦ Here a₀ = 0, a₁ = 0 and a₂ = 1
3. Some graphs of polynomial functions are shown below:

Fig.21.10

4. We can draw the graphs of polynomial functions without lifting the pen from the plane of the paper.
5. We can consider all polynomial functions as continuous functions.
• We will see detailed proof in the next section.

Solved example 21.14
Find all the points of discontinuity of the greatest integer function defined by f(x)=$\mathbf\small{\rm{\left\lfloor x\right\rfloor }}$, where $\mathbf\small{\rm{\left\lfloor x\right\rfloor }}$ denotes the greatest integer less than or equal to x.
Solution:
• Details about the greatest integer function can be seen in section 2.6.
• We must consider two cases:
♦ Case 1, where input x is not an integer.
♦ Case 2, where input x is an integer
These two cases will cover all real numbers.

Case 1: x not an integer.
1. Consider any arbitrary point c such that c is not an integer. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is:
$\mathbf\small{\rm{\left\lfloor c \right\rfloor }}$ = The greatest integer less than c.
• For example,
    ♦ if c = 3.58, then the limiting value = $\mathbf\small{\rm{\left\lfloor 3.58 \right\rfloor }}$ = 3
    ♦ if c = −1.5, then the limiting value = $\mathbf\small{\rm{\left\lfloor -1.5 \right\rfloor }}$ = −2
• It is clear that, $\lim_{x\rightarrow c} f(x)$ exists. So first condition is satisfied.
3. Applying condition (ii):
f(c) = $\mathbf\small{\rm{\left\lfloor c \right\rfloor }}$ = The greatest integer less than c.
• We see that: $\lim_{x\rightarrow c} f(x)~=~f(c)$
• So second condition is also satisfied.
4. Since both conditions are satisfied, the given function is continuous at x = c.
5. c was taken as an arbitrary point which is not an integer. So it can be any real number which is not an integer.
• Therefore we can write:
The given function is continuous at every real number which is not an integer.

Case 2: x is an integer
1. Consider any arbitrary point c such that c is an integer. Let us check whether the function is continuous at c.
2. Applying condition (i):
• Limiting value of f(x) at x = c is to be calculated.
We need to find the left side limit and right side limit.
    ♦ The left side limit will be (c-1).
    ♦ The right side limit will be c.
• For example,
    ♦ if c = 3, then the L.S.L = 2 and R.S.L = 3
    ♦ if c = −2, then the L.S.L = −3 and R.S.L = −2
(see graph in the fig.2.17 in section 2.6)
• It is clear that, $\lim_{x\rightarrow c} f(x)$ does not exist. So first condition is not satisfied.
3. We can establish continuity at a point only if both conditions are satisfied at that point.
4. c was taken as an arbitrary point which is an integer. So it can be any real number which is an integer.
• Therefore we can write:
The given function is not continuous at integers.

Conclusion:
• From cases 1 and 2, we see that,
♦ f(x) = is continuous at all real numbers which are not integers.
♦ But it is not continuous at every real number, which is an integer.

In the next section, we will see algebra of continuous functions.

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Tuesday, May 21, 2024

21.2 - Solved Examples on Continuity

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