With the Calculus as a key, Mathematics can be successfully applied to the explanation of the course of Nature – WHITEHEAD • 13.1 Introduction This chapter is an introduction to Calculus. Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. First, we give an intuitive idea of derivative (without actually defining it). Then we give a naive definition of limit and study some algebra of limits. Then we come back to a definition of derivative and study some algebra of derivatives. We also obtain derivatives of certain standard functions. 13.2 Intuitive Idea of Derivatives Physical experiments have confirmed that the body dropped from a tall cliff covers a distance of 4.9t2 metres in t seconds, i.e., distance s in metres covered by the body as a function of time t in seconds is given by s = 4.9t2. The adjoining Table 13.1 gives the distance travelled in metres at various intervals of time in seconds of a body dropped from a tall cliff. The objective is to find the veloctiy of the body at time t = 2 seconds from this data. One way to approach this problem is to find the average velocity for various intervals of time ending at t = 2 seconds and hope that these throw some light on the velocity at t = 2 seconds. Average velocity between t = t1 and t = t2 equals distance travelled between t = t1 and t = t2 seconds divided by (t2– t1). Hence the average velocity in the first two seconds Distance travelled between t2 = 2 and t 1 = 0 = Time interval ( t2 − t1) (19.6 − 0)m = = 9.8 ms(20)s /.− Similarly, the average velocity between t = 1 and t = 2 is (19.6 – 4.9 )m = 14.7 m/s21( −)s Likewise we compute the average velocitiy between t = t1 and t = 2 for various t1. The following Table 13.2 gives the average velocity (v), t = t1 seconds and t = 2 seconds. Table 13.2 Table 13.1 t s 0 0 1 4.9 1.5 11.025 1.8 15.876 1.9 17.689 1.95 18.63225 2 19.6 2.05 20.59225 2.1 21.609 2.2 23.716 2.5 30.625 3 44.1 4 78.4 t1 0 1 1.5 1.8 1.9 1.95 1.99 v 9.8 14.7 17.15 18.62 19.11 19.355 19.551 From Table 13.2, we observe that the average velocity is gradually increasing. As we make the time intervals ending at t = 2 smaller, we see that we get a better idea of the velocity at t = 2. Hoping that nothing really dramatic happens between 1.99 seconds and 2 seconds, we conclude that the average velocity at t = 2 seconds is just above 19.551m/s. This conclusion is somewhat strengthened by the following set of computation. Compute the average velocities for various time intervals starting at t = 2 seconds. As before the average velocity v between t = 2 seconds and t = t2 seconds is = Distance travelled between 2 seconds and 2 seconds t 2t 2− = Distance travelled in 2 secondst − Distance travelled in 2 seconds 2t 2− Distance travelled in t seconds− 19.6 = 2 t2 − 2 The following Table 13.3 gives the average velocity v in metres per second between t = 2 seconds and t2 seconds. Table 13.3 t2 4 3 2.5 2.2 2.1 2.05 2.01 v 29.4 24.5 22.05 20.58 20.09 19.845 19.649 Here again we note that if we take smaller time intervals starting at t = 2, we get better idea of the velocity at t = 2. In the first set of computations, what we have done is to find average velocities in increasing time intervals ending at t = 2 and then hope that nothing dramatic happens just before t = 2. In the second set of computations, we have found the average velocities decreasing in time intervals ending at t = 2 and then hope that nothing dramatic happens just after t = 2. Purely on the physical grounds, both these sequences of average velocities must approach a common limit. We can safely conclude that the velocity of the body at t = 2 is between 19.551m/s and 19.649 m/s. Technically, we say that the instantaneous velocity at t = 2 is between 19.551 m/s and 19.649 m/s. As is well-known, velocity is the rate of change of displacement. Hence what we have accomplished is the following. From the given data of distance covered at various time instants we have estimated the rate of change of the distance at a given instant of time. We say that the derivative of the distance function s = 4.9t2 at t = 2 is between 19.551 and 19.649. An alternate way of viewing this limiting process is shown in Fig 13.1. This is a plot of distance s of the body from the top of the cliff versus the time t elapsed. In the limit as the sequence of time intervals h1, h2, ..., approaches zero, the sequence of average velocities approaches the same limit as does the sequence of ratios Fig 13.1 CBCB CB1122 33 ,, ,... ACAC AC12 3 where C1B1 = s1 – s0 is the distance travelled by the body in the time interval h1 = AC1, etc. From the Fig 13.1 it is safe to conclude that this latter sequence approaches the slope of the tangent to the curve at point A. In other words, the instantaneous velocity v(t) of a body at time t = 2 is equal to the slope of the tangent of the curve s = 4.9t2 at t = 2. 13.3 Limits The above discussion clearly points towards the fact that we need to understand limiting process in greater clarity. We study a few illustrative examples to gain some familiarity with the concept of limits. Consider the function f(x)= x2. Observe that as x takes values very close to 0, the value of f(x) also moves towards 0 (See Fig 2.10 Chapter 2). We say lim ()=fx 0 x→0 (to be read as limit of f (x) as x tends to zero equals zero). The limit of f (x) as x tends to zero is to be thought of as the value f (x) should assume at x = 0. In general as x → a, f (x) → l, then l is called limit of the function f (x) which is lim fx lsymbolically written as → ()= . xa Consider the following function g(x)= |x|, x ≠0. Observe that g(0) is not defined. Computing the value of g(x) for values of x very near to 0, we see that the value of g(x) moves lim towards 0. So, x→0 g(x) = 0. This is intuitively clear from the graph of y = |x| for x ≠0. (See Fig 2.13, Chapter 2). Consider the following function. x −4hx = , x ≠2() 2 . x −2 Compute the value of h(x) for values of x very near to 2 (but not at 2). Convince yourself that all these values are near to 4. This is somewhat strengthened by considering the graph of the function y = h(x) given here (Fig 13.2). Fig 13.2 In all these illustrations the value which the function should assume at a given point x = a did not really depend on how is x tending to a. Note that there are essentially two ways x could approach a number a either from left or from right, i.e., all the values of x near a could be less than a or could be greater than a. This naturally leads to two limits – the right hand limit and the left hand limit. Right hand limit of a function f(x) is that value of f(x) which is dictated by the values of f(x) when x tends to a from the right. Similarly, the left hand limit. To illustrate this, consider the function ⎧1, x ≤0fx()=⎨ ⎩2, x >0 Graph of this function is shown in the Fig 13.3. It is clear that the value of f at 0 dictated by values of f(x) with x ≤ 0 equals 1, i.e., the left hand limit of f (x) at 0 is lim fx() =1 . x→0 Similarly, the value of f at 0 dictated by values of f (x) with x > 0 equals 2, i.e., the right hand limit of f (x) at 0 is lim fx() =2 →+ . x 0 Fig 13.3 In this case the right and left hand limits are different, and hence we say that the limit of f (x) as x tends to zero does not exist (even though the function is defined at 0). SummaryWe say lim f(x) is the expected value of f at x = a given the values of f near x→a – x to the left of a. This value is called the left hand limit of f at a. We say lim f () x is the expected value of f at x = a given the values of→+xa f near x to the right of a. This value is called the right hand limit of f(x) at a. If the right and left hand limits coincide, we call that common value as the limit of f(x) at x = a and denote it by lim f(x).x→a Illustration 1 Consider the function f(x) = x + 10. We want to find the limit of this function at x = 5. Let us compute the value of the function f(x) for x very near to 5. Some of the points near and to the left of 5 are 4.9, 4.95, 4.99, 4.995. . ., etc. Values of the function at these points are tabulated below. Similarly, the real number 5.001, 5.01, 5.1 are also points near and to the right of 5. Values of the function at these points are also given in the Table 13.4. Table 13.4 x 4.9 4.95 4.99 4.995 5.001 5.01 5.1 f(x) 14.9 14.95 14.99 14.995 15.001 15.01 15.1 From the Table 13.4, we deduce that value of f(x) at x = 5 should be greater than 14.995 and less than 15.001 assuming nothing dramatic happens between x = 4.995 and 5.001. It is reasonable to assume that the value of the f(x) at x = 5 as dictated by the numbers to the left of 5 is 15, i.e., lim ()=fx 15 .5– x→ Similarly, when x approaches 5 from the right, f(x) should be taking value 15, i.e., lim ()=fx 15 →+ . x 5 Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to 15. Thus, () ()() .lim fx =lim fx =lim fx =15 x→5− x→5+ x→5 This conclusion about the limit being equal to 15 is somewhat strengthened by seeing the graph of this function which is given in Fig 2.16, Chapter 2. In this figure, we note that as x approaches 5 from either right or left, the graph of the function f(x) = x +10 approaches the point (5, 15). We observe that the value of the function at x = 5 also happens to be equal to 15. Illustration 2 Consider the function f(x) = x3. Let us try to find the limit of this function at x = 1. Proceeding as in the previous case, we tabulate the value of f(x) at x near 1. This is given in the Table 13.5. Table 13.5 From this table, we deduce that value of f(x) at x = 1 should be greater than 0.997002999 and less than 1.003003001 assuming nothing dramatic happens between x = 0.999 and 1.001. It is reasonable to assume that the value of the f(x) at x = 1 as dictated by the numbers to the left of 1 is 1, i.e., lim ()= .fx 1 x→1− Similarly, when x approaches 1 from the right, f(x) should be taking value 1, i.e., lim ()= .fx 1 →+x 1 Hence, it is likely that the left hand limit of f(x) and the right hand limit of f(x) are both equal to 1. Thus, lim ()= f (x)=lim fx lim f (x)=1. x→1− x→1+ x→1 This conclusion about the limit being equal to 1 is somewhat strengthened by seeing the graph of this function which is given in Fig 2.11, Chapter 2. In this figure, we note that as x approaches 1 from either right or left, the graph of the function f(x) = x3 approaches the point (1, 1). We observe, again, that the value of the function at x = 1 also happens to be equal to 1. Illustration 3 Consider the function f(x) = 3x. Let us try to find the limit of this function at x = 2. The following Table 13.6 is now self-explanatory. Table 13.6 x 1.9 1.95 1.99 1.999 2.001 2.01 2.1 f(x) 5.7 5.85 5.97 5.997 6.003 6.03 6.3 As before we observe that as x approaches 2 from either left or right, the value of f(x) seem to approach 6. We record this as fx lim x ()lim ()= f () =lim fx =6 x→2− x→2+ x→2 Its graph shown in Fig 13.4 strengthens this fact. Here again we note that the value of the function at x = 2 coincides with the limit at x = 2. Illustration 4 Consider the constant function f(x) = 3. Let us try to find its limit at x = 2. This function being the constant function takes the same Fig 13.4 value (3, in this case) everywhere, i.e., its value at points close to 2 is 3. Hence lim fx =lim fx =lim fx =3() ()()x→2 x→2+ x→2 Graph of f(x) = 3 is anyway the line parallel to x-axis passing through (0, 3) and is shown in Fig 2.9, Chapter 2. From this also it is clear that the required limit is 3. In fact, it is easily observed that lim xa→ ()fx 3= for any real number a. Illustration 5 Consider the function f(x) = x2 + x. We want to find 1 lim x→ f ()x . We tabulate the values of f(x) near x = 1 in Table 13.7. Table 13.7 x 0.9 0.99 0.999 1.01 1.1 1.2 f(x) 1.71 1.9701 1.997001 2.0301 2.31 2.64 From this it is reasonable to deduce that lim ()=lim f ()=lim fxfx x ()=2 x→1− x→1+ x→1. From the graph of f(x) = x2 + x shown in the Fig 13.5, it is clear that as x approaches 1, the graph approaches (1, 2). Here, again we observe that the lim x→1 f (x) = f (1) Now, convince yourself of the Fig 13.5following three facts: lim x2 =1, lim x =1 and lim x 1+= 2 x→1 x→1 x→1 2 ⎡2 ⎤Then lim x + lim x 11 2 =lim x +x .=+= x→1 x→1 x→1 ⎣⎦Also lim x. lim (x +=1) 1.2 =2 =lim ⎡x(x +1)⎤=lim ⎡x2 +x⎤.⎣ ⎦⎣⎦x→1 x→1 x→1 x→1 Illustration 6 Consider the function f(x) = sin x.We are interested in lim sin x,π 2 x→ where the angle is measured in radians. Here, we tabulate the (approximate) value of f(x) near 2 π (Table 13.8). From this, we may deduce that 2 lim x −π→ ()fx= 2 lim x +π→ ()fx= 2 lim x π→ () 1fx= . Further, this is supported by the graph of f(x) =sin xwhich is given in the Fig 3.8 (Chapter 3). In this case too, we observe that lim sin x= 1. π x→2 Table 13.8 x 0.1 2 π− 0.01 2 π− 0.01 2 π+ 0.1 2 π+ f(x) 0.9950 0.9999 0.9999 0.9950 Illustration 7 Consider the function f(x)= x+ cos x. We want to find the lim f(x). x→0 Here we tabulate the (approximate) value of f(x) near 0 (Table 13.9). Table 13.9 x – 0.1 – 0.01 – 0.001 0.001 0.01 0.1 f(x) 0.9850 0.98995 0.9989995 1.0009995 1.00995 1.0950 From the Table 13.9, we may deduce that lim ()= fx ()=lim fx lim fx()=1 x→0− x→0+ x→0 In this case too, we observe that lim f(x) = f(0) = 1.x→0 Now, can you convince yourself that lim [x+cos x]=lim x+lim cos xis indeed true? x→0 x→0 x→0 fx=Illustration 8 Consider the function () 1 for > . We want to know lim f(x).2 x 0 x→0x Here, observe that the domain of the function is given to be all positive real numbers. Hence, when we tabulate the values of f(x),it does not make sense to talk of xapproaching 0 from the left. Below we tabulate the values of the function for positive xclose to 0 (in this table ndenotes any positive integer). From the Table 13.10 given below, we see that as xtends to 0, f(x) becomes larger and larger. What we mean here is that the value of f(x) may be made larger than any given number. Table 13.10 x 1 0.1 0.01 10–n f(x) 1 100 10000 102n Mathematically, we say lim fx=+∞()x→0 We also remark that we will not come across such limits in this course. lim fxIllustration 9 We want to find (), wherex→0 ⎧x−2, x<0 ()=⎨⎪0, xfx =0 ⎪x+2, x>0⎩ As usual we make a table of xnear 0 with f(x).Observe that for negative values of x we need to evaluate x– 2 and for positive values, we need to evaluate x+ 2. Table 13.11 x – 0.1 – 0.01 – 0.001 0.001 0.01 0.1 f(x) – 2.1 – 2.01 – 2.001 2.001 2.01 2.1 From the first three entries of the Table 13.11, we deduce that the value of the function is decreasing to –2 and hence. lim fx=−2 →− ()x 0 From the last three entires of the table we deduce that the value of the function is increasing from 2 and hence lim fx=2 →+x 0 () Since the left and right hand limits at 0 do not coincide, we say that the limit of the function at 0 does not exist.Graph of this function is given in the Fig13.6. Here, we remark that the value of the function at x= 0 is well defined and is, indeed, equal to 0, but the limit of the function at x= 0 is not even defined. Fig 13.6lim fxIllustration 10 As a final illustration, we find (),x→1 where ⎧x+2 x≠1fx=() ⎨ 0 x=1⎩ Table 13.12 x 0.9 0.99 0.999 1.001 1.01 1.1 f(x) 2.9 2.99 2.999 3.001 3.01 3.1 As usual we tabulate the values of f(x) for xnear 1. From the values of f(x) for xless than 1, it seems that the function should take value 3 at x= 1., i.e., () .lim fx=3 x→1− Similarly, the value of f(x) should be 3 as dictated by values of f(x) at xgreater than 1. i.e. lim fx=3 x→1+ () . But then the left and right hand limits coincide and hence Graph of function given in Fig 13.7 strengthens our deduction about the limit. Here, we Fig 13.7 lim fx=lim fx=lim fx =3() ()() . x→1− x→1+ x→1 note that in general, at a given point the value of the function and its limit may be different (even when both are defined). 13.3.1 Algebra of limits In the above illustrations, we have observed that the limiting process respects addition, subtraction, multiplication and division as long as the limits and functions under consideration are well defined. This is not a coincidence. In fact, below we formalise these as a theorem without proof. Theorem 1 Let fand gbe two functions such that both lim f(x) and lim g(x) exist. x→ax→a Then (i) Limit of sum of two functions is sum of the limits of the functions, i.e., lim [f(x) + g(x)]= lim f(x) + lim g(x).x→a x→ax→a (ii) Limit of difference of two functions is difference of the limits of the functions, i.e., lim [f(x)– g(x)] = lim f(x) – lim g(x). x→a x→ax→a (iii) Limit of product of two functions is product of the limits of the functions, i.e., lim [f(x). g(x)]= lim f(x). lim g(x). x→a x→ax→a (iv) Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e., lim f()fxlim ()=xa→ x xa x lim xa → g() → g()x $Note In particular as a special case of (iii), when gis the constant function such that g(x)= λ,for some real number λ, we have lim ⎡(λ. f)(x)⎤=λ.lim f(x)⎣ ⎦→ . xa→ xa In the next two subsections, we illustrate how to exploit this theorem to evaluate limits of special types of functions. 13.3.2 Limits of polynomials and rational functions A function fis said to be a npolynomial function if f(x) is zero function or if f(x)= a+ a+ a2 +. . . + ax,01x2xnwhere aisare real numbers such that an≠ 0 for some natural number n. We know that lim x= a. Hence x→a lim x2 =lim (xx. )=lim x.lim = . aa 2xa= → xa xa xa xa→ →→ An easy exercise in induction on ntells us that lim xn=an xa→ Now, let f()= xa+ 2 ++ axnbe a polynomial function. Thinking xaa+ x ...01 2 n of each of , xax2,..., xnas a function, we haveaa, a01 2 n lim ()= ⎡0 + 1xa2 x2 ++ ... nnfx lim aa+ ax⎤ xa → →⎣xa ⎦ a 1 a2 ++ lim an n = lim 0 +lim ax+lim 2 x ... x xa→ xa → xa→ xa → = 0 + lim + 2 lim x2 ... alim naa 1 xa ++ nx xa→ xa→ xa→ = aa+ aaa++ ... aa01 + 22 nn = f(a) (Make sure that you understand the justification for each step in the above!) x A function fis said to be a rational function, if f(x)= hx, where g(x) and h(x) g() () are polynomials such that h(x) ≠ 0. Then lim ()= lim gx()=gxxa a() → g()fx lim = xa→ xa hx hh()xa → () lim ()xa → However, if h(a) = 0, there are two scenarios – (i) when g(a) ≠ 0 and (ii) when g(a) = 0. In the former case we say that the limit does not exist. In the latter case we )kcan write g(x)= (x– ag1 (x), where kis the maximum of powers of (x– a) in g(x) )lSimilarly, h(x)= (x– ah1(x) as h(a) = 0. Now, if k> l,we have ()x lim (xagx () lim g−)k 1 xa→ xalim () =→ → fx= lim () lxa xahx lim (−) () xahx 1→ xa→ −( ) ()lim (xa−)klgx gaxa 1 0. ()→ 1==0= lim hx → () ha ()11 xa If k< l,the limit is not defined. ⎣ 32 ⎤⎦⎣() ⎦Example 1 Find the limits: (i) lim ⎡x−x+1 (ii) lim ⎡xx+1 ⎤ x→1 x→3 ⎡ 2 10 ⎤(iii) lim 1 xx++ ...++ x .⎣ ⎦x→−1 Solution The required limits are all limits of some polynomial functions. Hence the limits are the values of the function at the prescribed points. We have lim 3(i) x→1[x– x2 + 1] = 13 – 12 + 1 = 1 lim ⎡xx+1 ⎤=33 1 34 =12 += (ii) ⎣( )⎦( )()x→3 210 210 (iii) lim 1 xx++ ... x⎦⎤ 1 ()() 1 + +− 1⎡++⎣ =+− +− ... ()1 x→−1 1 1 1... +=1 1.=−+ Example 2 Find the limits: 2 32⎡x+1 ⎤ ⎡x−4x+4x⎤ (i) lim ⎢⎥ (ii) lim ⎢ 2 ⎥ x→1 ⎣x+100 x→2 x−4⎦ ⎣⎦ 2 32⎡ x−4 ⎤ ⎡x−2x ⎤ (iii) lim ⎢32 ⎥ (iv) lim ⎢2 ⎥ x→2 x−4x+4xx→2 x−5x+6⎣⎦ ⎣⎦ ⎡x−21 ⎤ (v) lim ⎢2 −32 ⎥. x→1 ⎣x−xx −3x+2x⎦Solution All the functions under consideration are rational functions. Hence, we first 0evaluate these functions at the prescribed points. If this is of the form , we try to0 rewrite the function cancelling the factors which are causing the limit to be of 0the form 0. x2 +1 12 +12(i) We have lim == x→1 x+100 1 +100 101 0(ii) Evaluating the function at 2, it is of the form 0. 32x ( 2 −4x+4x xx−2)Hence lim 2 = lim x→2 x−4 x→2 (x+2)( x−2) (−)xx 2lim as x≠2= x→2 (x+2) () 022 −2 = ==0.22 4+ 0(iii) Evaluating the function at 2, we get it of the form 0. 2 (x+2)(x−2)x−4 lim Hence lim = 232 x→2−(x→2 x 4x+4x xx−2)(x+2)+ 422lim == = x 2 (−2) 22 (−2) 0→ xx which is not defined. 0(iv) Evaluating the function at 2, we get it of the form 0. x3 −2x2 lim xx2 (−2)Hence lim = x→2 x2 −5x+6 x→2 (x−2)( x−3) 2x ()22 4 = lim = ==−4. x→2 (x−3) 2 −3 −1 (v) First, we rewrite the function as a rational function. ⎡⎤ ⎡ x− 2 − 1 ⎤⎢ x− 21 ⎥−⎢ 2 32 ⎥ = ⎢ xx 2 ⎥⎣ x− xx − 3x+ 2x⎦(−1) xx(− 3x+ 2)⎣⎦ ⎡ x− 21 ⎤ ⎢− ⎥= ⎢⎣ ( 1) −)( x− 2 ⎥⎦xx− xx( 1 ) ⎡ x2 − 4x+ 4 −1 ⎤ ⎢⎥= ⎢(−1)( x− 2)⎥xx⎣⎦ x2 − 4x+ 3 = (−1)( x− 2)xx 0Evaluating the function at 1, we get it of the form 0. ⎡ x2 − 21 ⎤ x2 − 4x+ 3lim − lim Hence x→1 ⎢⎣ x2 − xx 3 − 3x2 + 2x ⎥⎦ = x→1 xx(−1)( x− 2) (x− 3)( x−1) = x→1 lim (−1)( x− 2)xx x− 3 13− = x 1 = = 2.→ 2 11lim xx(−) (−2) We remark that we could cancel the term (x– 1) in the above evaluation because x≠ 1. Evaluation of an important limit which will be used in the sequel is given as a theorem below. Theorem 2 For any positive integer n, xn− an n−1lim = na . xa→ x− a Remark The expression in the above theorem for the limit is true even if nis any rational number and ais positive. Proof Dividing (xn – an) by (x – a), we see that xn – an = (x–a) (xn–1 + xn–2 a + xn–3 a2 + ... + x an–2 + an–1) xn −an Thus, lim =lim (xn–1 + xn–2 a + xn–3 a2 + ... + x an–2 + an–1)x→ax −ax→a = an – l + a an–2 +. . . + an–2 (a) +an–l = an–1 + an – 1 +...+an–1 + an–1 (n terms) n−1 = na Example 3 Evaluate: x15 −1(i) lim 10x→1 x −1 Solution (i) We have x15 −1lim = x→1 x10 −1 = = = (ii) Put y = 1 + x, so that 1 x 1+−Then lim = x→0 x = = 1 x 1+−(ii) lim x→0 x 15 10⎡x −1 x −1⎤lim ÷⎢⎥ x→1 x −1 x −1⎣⎦ 15 10 ⎡x −1⎤⎡x −1⎤lim ÷lim ⎢⎥⎢ ⎥ x→1 x −1 x→1 x −1⎣⎦⎣ ⎦ 15 (1)14 ÷ 10(1)9 (by the theorem above) 315 ÷ 10 = 2 y →1 as x →0. y −1lim y→1 y –1 11 y 2 −12 lim y→1 y −1 1 −1112(1) (by the remark above) = 22 13.4 Limits of Trigonometric Functions The following facts (stated as theorems) about functions in general come in handy in calculating limits of some trigonometric functions. Theorem 3 Let f and g be two real valued functions with the same domain such that f (x) ≤ g( x) for all x in the domain of definition, For some a, if both lim f(x) and x→alim g(x) exist, then lim f(x) ≤ lim g(x). This is illustrated in Fig 13.8.x→ax→ax→aTheorem 4 (Sandwich Theorem) Let f, g and h be real functions such that f(x) ≤ g( x) ≤ h(x) for all x in the common domain of definition. For some real number a, if lim f(x)= l = lim h(x), then lim g(x)= l. This is illustrated in Fig 13.9.x→ax→ax→aGiven below is a beautiful geometric proof of the following important inequality relating trigonometric functions. sin x π cos x <<1 for 0 < x < (*)x 2 Proof We know that sin (– x) = – sin x and cos( – x) = cos x. Hence, it is sufficient xto prove the inequality for 0 <<π .2 In the Fig 13.10, O is the centre of the unit circle such that πthe angle AOC is x radians and 0 < x < . Line segments B A and 2 CD are perpendiculars to OA. Further, join AC. Then Area of ΔOAC< Area of sector OAC < Area of ΔOAB. Fig 13.10 1 x 21i.e., OA.CD < .π.(OA)