FUNCTIONS ,LIMITS AND CONTINUTY
PREPARED BY
AJAY BENIWAL
Real Functions and their Graphs
Functions
f is a function from set A to a set B if each element x in A can be associated with a unique element in B.
The unique element B which f associates with x in A denoted by f (x).
Domain
In the above definition of the function, set A is called domain.
Co-Domain
In the above definition of the function, set B is called co-domain.
Real Function
A real valued function f : A to B or simply a real function 'f ' is a rule which associates to each possible real number xA, a unique real number f(x)B, when A and B are subsets of R, the set of real numbers.
In other words, functions whose domain and co-domain are subsets of R, the set of real numbers, are called real valued functions.
Value of a Function
If 'f ' is a function and x is an element in the domain of f, then image
f(x) of x under f is called the value of 'f ' at x.
Types of Function and their Graphs
Constant function
A function f : A � B Such that A, B � R, is said to be a constant function if there exist K � B such that f(x) = k.
Domain = A
Range = {k}
The graph of this function is a line or line segment parallel to x-axis. Note that, if k>0, the graph B is above X-axis. If k<0, the graph is below the x-axis. If k = 0, the graph is x-axis itself.
Identity function
A function f : R� R is said to be an identity function if for all x �R, f(x) = x.
Domain = R
Range = R
Polynomial function
A function f : R� R is said to be a polynomial function if for each x� R, f(x) is a polynomial in x.
f(x) = x3 + x2 + x
Modulus function
f : R� R such that f(x) = |x|, is called the modulus function or absolute value function.
Domain = R
Square root function
Since square root of a negative number is not real, we define a function f : R+ �R such that
Greatest integer function or Step function (Floor function)
f (x) = [x] = greatest integer less than or equal to x
[x] = n, where n is an integer such that
Smallest integer function (Ceiling function)
For a real number x, we denote by [x], the smallest integer greater than or equal to x. For example, [5 . 2] = 6, [-5 . 2] = -5, etc. The function f:RR defined by
f(x) = [x], xR
is called the smallest integer function or the ceiling function.
Domain: R
Range : Z
Exponential function
The exponential function is defined as f(x) = ex. Its graph is
Logarithmic function
Logarithmic function is f (x) = log x. Its graph is
Trigonometric functions
Trigonometric functions are sinx, cosx, tanx, etc. The graph of these functions have been done in class XI.
Inverse functions
Inverse functions are sin-1x, cos-1x, tan-1x etc. The graph of these functions have been done in class XI.
Signum functions
Odd function
A function f : AB is said to be an odd function if
f(x) = - f(-x) for all xA
The domain and range of f depends on the definition of the function.
Examples of odd function are
y = sinx, y = x3, y = tanx
Even function
A function f : AB is said to be an even function if
f(x) = f(-x) for all xA.
The domain and range of f depends on the definition of the function.
Examples of even function are
y = cosx, y = x2, y = secx
A polynominal with only even powers of x is an even function.
Reciprocal function
Operation on Real Functions
Functions can be added, subtracted and multiplied. They can also be divided where the divisor function does not take the value zero. These operations create new functions.
If f(x) and g(x) are two real valued functions, then for every value of x that belongs to both the domains of f and g, we can define the following functions:
Sum Function: (f + g) (x) = f (x) + g (x)
Difference Function: (f - g) (x) = f(x) - g(x)
Product Function: fg(x) = f(x) g(x)
Domain of these functions are {Domain f} {Domain g}
Quotient Function
Scalar Multiplication Function
(c f) (x) = c.f (x) for all x Domain (f)
Example:
Let f(x) =
g(x) =
{Domain (f) = (0,)
f + g (x) = f(x) + g (x) Domain
[0,1]
[0,1]
[0,1]
Composite Functions
If range(f) C Dom(g), we define the composite function of g and f (gof) by
gof(x) = g [f(x)] for all xA
If range (g) C dom f, we define the composite function (fog) of f and g by
fog (x) = f [g(x)] for all xX
Example:
Let f(x) = x2-1, g(x) = 3x-1
Domain f = R
Domain g = R
fog (x) = f [g(x)]
(range of g C domain of f )
= f(3x-1)
= (3x-1)2 - 1
= 9x2 + 1 - 6x - 1
= 9x2 - 6x
Domain fog = R
Inverse Functions
Let f : AB be a real valued one-one and onto function.
Therefore, we can define a function, (denoted by f-1) called 'the inverse of f ' as follows:
f-1 : BA such that
Example:
If f: R R is defined by
f(x) = 5x - 7, find f-1(x) and f-1(8).
Suggested answer:
Let f (x) = y, then y = 5x - 7
f is a one-one and onto function.
This is inverse function of f.
Limits But we are interested in finding the value of y near 2. When x = 1.9, y = 3.9 x = 1.99, y = 3.99 x = 1.999 y = 3.999 . . . . . . Similarly, when x = 2.1 y = 4.1 x = 2.01 y = 4.01 x = 2.001 y = 4.001 . . . . . . Observe that as x approaches 2, y approaches 4. 4 is called the limit of f(x) as x approaches 2 and is represented as If x is close to 2, f(x) is close to 4. If |x - 2| is small then |f(x) - 4| is also small. Note 2: When x approaches 2 from left, the limit obtained is called left hand limit and when x approaches 2 from the right, the limit obtained is called the right hand limit. Note 3: We say that because both left hand limit and right hand limit are finite and equal. If there exists a real number l such that if |f (x) - l| can be made as small as we possible by taking x sufficiently close to a, then l is called the limit of f (x) as x tends to 'a'. Left Hand Limit Let f(x) tend to a limit l1 as x tends to a through values less than 'a', then l1 is called the left hand limit which is symbolically written as Right Hand Limit Let f(x) tend to a limit l2 as x tends to 'a' through values greater than 'a', then l2 is called the right hand limit which is symbolically written as Note: We say that limit of f(x) exists at x = a, if l1 and l2 are both finite and equal. Some Properties of Limits Limit of composite functions Limits of Polynomial Functions and Rational Functions a) Limits of polynomial functions can be found by substitution If f(x) = anxn + an-1xn-1+an-2xn-2+��.. a0, then Example: b) Limit of a rational function can be found by substitution, if the limit of the denominator is not zero. In other words, where P(x) and Q(x) are polynomials, then we have Example: c) If Q(c)=0, common factors between the numerator and denominator are identified and cancelled. This reduces the rational function to another rational function whose denominator is not zero at c. Example: If Q(c) = 0, we can also factorise and cancel a common factor. Example: Suggested answer: d) In the rational function, we have If P(c) = a, Q(c ) = 0, then Example: At x =1, the numerator of the rational function 6, but denominator is 0. Theorem 1: Proof: Case (i): If n is a positive integer. We have, Case (ii): If n is a negative integer. Let n = -k, where k is a positive integer Hence true for negative integers. Case (iii): When n is a rational number. Hence the theorem.
Limits (Contd....) Limits of Trigonometric Functions Before describing the limits of trigonometric functions, we state few theorems along with Sandwich theorem, which helps in calculating a variety of limits in subsequent chapters. Theorem 2: Let f and g be real valued functions defined on an interval containing c such that exist. Then The following statement is not true. f(x) < g(x) for all x Theorem 3: If f is a function defined on an open interval containing c, then Theorem 4 (Sandwich Theorem): for all x in some open interval containing c and suppose Since f is sandwiched between two functions g and h, the above theorem is known as sandwich theorem. Theorem 5: Proof: Consider a circle with centre O and radius r. Join AB. Let the tangent at B meet OA produced at P. Draw BN perpendicular to OA. From ONB, BN = r sin From OBP, BP = r tan From the figure, we have Area of triangle OAB < Area of sector OAB < Area of triangle OBP Note 1: Note 2: = 1 Note 3: Limits Involving Exponential Functions Theorem 6: Proof: We know that, Further, we have Substituting this value in (2), we have From (1) and (3), we have it follows from the above inequation that From equation(4), we get (Note that -x > 0, so multiplying this in equation by -x, the inequality remains same) Add 1 on both sides, Taking the reciprocal, we have Subtracting 1, we have Diving by the negative number x, we get Now x < 0, |x| = - x From (5) and (6), the theorem is proved. Theorem 7: Proof: From the above theorem, we have = 1 Using Sandwich theorem, we get Example: Suggested answer: =1 Theorem 8: Proof: Theorem 9: Proof:
Limits (Contd....) Limits at Infinity and Infinite Limits Limits at infinity If x is a variable such that it can take any real value how much ever If x is a variable such that it can take any real value how much ever Example: We write Infinite limits Let f(x) be a function of x, if the value of f(x) can be made greater than any pre-assigned number by taking x close to 'a', then we say Similarly, if the value of f (x) can be made less than any pre-assigned number by taking x close to 'a', then we say f (x) tends to - as x approaches 'a' Example: In the first quadrant, as x approaches zero from the right of zero, the value of f(x) increases without bound. It becomes greater than any assigned positive real value. In this case, we say or Similarly, in the third quadrant as x approaches zero from the left, the value of decreases without bound. It can be smaller than any pre-assigned negative real number. The following statement is useful to evaluate limits at infinity of rational functions. Example: Evaluate the limit: Suggested answer: One Sided Limit We have discussed earlier about right hand limit and left hand limit. Both these limits are called one sided limits. x approaches a from the right side and through values greater than a. For a function f(x), we say as left hand Limit, as x approaches a from the left and through the values lesser than a. The two important properties of these one-sided limits that i) If the left hand limit and right hand limit of a function at a point exists, but are not equal, then we conclude that the limit at that point does not exist. ii) If LHL and RHL of a function at a point (say a) exist and they are equal, we conclude that limit at that point exists and we write We conclude our discussion on limits with one example on one sided limits. Example: For the function f(x) = x+ (x-[x])2, find the RHL and LHL at x = 2. Check whether Suggested answer: RHL = = 2 = 3
Continuity at a Point A function f (x) is said to be continuous at x = a if f (a) exists. Continuity in an Interval A function f(x) is said to be continuous in an interval I, if it is continuous at each point of I. then x can approach c both from the left and the right and so for f (x) to be continuous at c, we must have If the interval I is the closed interval [a,b], then x cannot approach a from the left and it cannot approach b from the right. In this case f (x) is continuous at a, if it is continuous at b, if and it is continuous at Example: Discuss the continuity of the function f given by f(x) = |x - 1| + |x - 2| at x=1 Suggested answer: Right hand limit at x = 1 = h + (-h+1) = 1 Left hand limit at x = 1 =1 f(1) = |1 - 1| + |1 - 2| = 1 The function is continuous at x = 1. Note 1: We say that f(x) is continuous if f(x) is continuous at every point in its domain. Note 2: If f and g are two continuous functions then f + g, f - g, fg are continuous functions. Note 3: Every polynomial function is continuous. Note 4: Every rational function is continuous at each point of its domain. Note 5: Composition of two continuous functions is continuous. Removal Discontinuity of a Function at a Point If limit of a function f exists at a point c, but it is not equal to the value of the function at c. But if f(c) l, then the function is said to have removal discontinuity at x = c. This type of function can be made continuous by charging the value of f (c). If we change f(c) = l, the function becomes continuous at x = c. Example: Show that the function has a removal discontinuity at x = 4. Redefine the function f(x) at x = 4 to make it continuous. Suggested answer: f (x) is not continuous at x = 4. If we define f (x) = 256 at x = 4, then Theorem 10: If f and g are real functions such that fog is defined, if g is continuous at a point c, and if f is continuous at g(c), then fog is continuous at c. Proof: Since g is continuous at c, we have Again, f is continuous at g(c) and so Hence fog is continuous at c.
Summary Algebra of limits Step I: Factor of f(x) and g(x). Step II: Common factors of numerator and denominator are Step III: Substitute x = a and obtain the limits. Standard Results Limits at infinity and infinite limits: Infinite limit of a function: If f(x) can be made as small as possible (negatively infinite) as x � a we say A real function f(x) is said to be continuous at x = a, if (i) f(a) is defined If f1 and f2 are continuous functions, then Every polynomial is continuous. Every rational function is continuous. Conclusion In this chapter, we have studied various types of functions and their graphs. The use of graphs also facilitate the study of domain and range of functions. We have studied the methods of evaluating limits of function of various kinds which is very useful iQuestion Bank
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1. Find the domain and range of each of the following functions:
Answer:
Now, f (x) = x 2 - 6x + 5 = (x - 3) 2 - 4 iii) h (x) = 1 - |x| h(x) is defined for all x in R. Domain of h (x) is R g(x) is defined for all x in R. Domain of g is R.
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2. Find the domain and range of the following functions:
Answer:
i) Domain of cos x is R
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3. Draw the graph of the function
Answer:
f (x) is a
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4. and indicate the domain of each function.
Answer:
(gof)(x) = x
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5. Evaluate:
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6. Evaluate:
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7. Evaluate:
Answer:
Divide each term in numerator and denominator by x,
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8. Evaluate:
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9. Evaluate:
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10. Evaluate:
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11. Evaluate:
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12. Evaluate:
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13. Evaluate:
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14. Evaluate:
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= (cos a) x (1) = cos a
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15. Evaluate:
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16. Evaluate:
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= e 2
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17. Evaluate:
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= 1 - 1 = 0
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18. Evaluate:
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Divide both numerator and denominator by x 2 ,
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19. Evaluate:
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20. Evaluate:
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21. Evaluate:
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Let x = sin
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22. Evaluate:
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23. Evaluate:
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24. Evaluate:
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Dividing each term in numerator and denominator by x, then we get
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25. Evaluate:
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26. Evaluate:
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27.
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28. Evaluate:
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29. Evaluate:
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= 1
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30. Evaluate:
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31.
Answer:
The function can be redefined as The function is discontinuous.
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32. A function f (x) is defined as
Answer:
f (3) = 5
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33. Find the point of discontinuity of the function
Answer:
= 18 x 6 = 108 f (3) = 81
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34. For what value of k is the following function continuous at x = 0.
Answer:
By definition of continuity, we have k = 1
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35.
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36.
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37.
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38.
Answer:
We have f (0) = 0
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39.
Answer:
f (0) = 0
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40. then find the value of a, b and c, such that f (x) is continuous at x = 0.
Answer:
f(0) = c Since f(x) is continuous, we have LHL = f(0) = RHL c = 1/2
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41. Find the domain and the range of each of the following functions: (i) f(x) = 1 - |x - 3| (ii) f(x) = x!
Answer:
(i) Domain - R, the set of real numbers (ii) Domain = the set of whole numbers i.e., (0, 1, 2.........) Range = {n! : n = 0, 1, 2 .........}
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42. Draw the graph of each of the following functions: (i) f(x) = 1/x (ii) f(x) = |x - 2| + |x - 3| (iv) f(x) = x!
Answer:
If x is positive, f(x) is positive and if x is negative, f(x) is negative. The graph is as given below. (ii) f(x) = |x - 2|+|x - 3| = 5 - 2x The graph is as given below. (iii) f(x) = 3 - x if x > 1 = 1 if x = 1 =2x if x < 1 (iv) f(x) = x! The domain is the set of whole numbers. f(0) = 1 f(1) = 1 f(2) = 2 f(3) = 6 etc. The graph is as given below.
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43. Find fog and gof if
Answer:
gof(x) = g(x+1) = e x + 1 (ii) fog(x) = f(sinx 2 ) = gof(x) = g( ) = sin 2 gof(x) = g(x 2 + 2)
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44. Evaluate the following limits:
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45. Evaluate the following limits:
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46.
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47. Discuss the continuity of the function. at x = 3.
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The function is discontinous at x = 3.
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48. If
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As the function is continuous,
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49. Find the value of a and b so that the function ' f ' given by is continuous at x = 3 and x = 5.
Answer:
f(3) = 1 As the function is continuous at x = 3, 1 = 3a + b ...(1) f(5) = 7 As the function is continous at x = 5, we have 5a + b = 7 ...(2) 3a + b = 1 5a + b = 7 -2a = -6 b = -8
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50. Show that the function 'f ' given by f(x) = |x|+|x - 1|, x R is continuous both at x = 0 and x = 1.
Answer:
f(0) = |0|+ |0 - 1| = 1 f(1) = |1| + |1 - 1| = 1
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51. Find the domain and range of the following functions: (i) f(x) = sgn(x - 2) (ii) f(x) = x 2 - [x 2 ]
Answer:
(i) Domain : R Range : {-1, 0, 1} ( sgn (x - 2) = 1 if x > 2 = 0 if x = 2 = -1 if x < 2) (ii) Domain = R Range = {0, 1} (iii) Domain = R - {4} Range = {-1}
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52.
Answer:
As f(x) is an odd function, f(-x) = - f(x) - x 3 - Kx 2 - 2x = -[x 3 - Kx 2 + 2x] -2Kx 2 = 0 K = 0
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53.
Answer:
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54. Let 'f ' be a function such that f(x + y) = f(x) + f(y), x, y R show that if 'f ' is continuous at x = 0, then it is continuous everywhere.
Answer:
f(x + y) = f(x) + f(y) Let x = y = 0 f(0) = f(0) + f(0) As 'f ' is continuous at x = 0, we have Let x be any value R. = f(x) + 0 = f(x) f(x) is continuous everywhere.
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55. Discuss the continuity of the function 'f ' given by f(x) = |x - 1| + |x - 2| at x = 1 and x = 2.
Answer:
f(x) = |x - 1| + |x - 2| f(1) = |1 - 1| + |1 - 2| = 1 f(x) is continuous at x = 1. f(2) = |2 - 1| + |2 - 2| = 1 f(x) is continuous at x = 2.
n the study of differential calculus.