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Hence theright-hand side of the last inequality (and therefore the left-hand side also) tendsto A.Example 19 limx→0 sinx x = limx→0 cos1 x = 1.This example should not be looked on as a new, independent proof of the relationsin xx → 1 as x → 0. The fact is that in deriving the relation sin x = cos x we alreadymade use of the limit just calculated.We always verify the legitimacy of applying l’Hôpital’s rule after we find thelimit of the ratio of the derivatives. In doing so, one must not forget to verify condition 10 or 20 .
The importance of these conditions can be seen in the followingexample.2505Differential CalculusExample 20 Let f (x) = cos x, g(x) = sin x. Then f (x) = − sin x, g (x) = cos x, (x)(x)and fg(x)→ +∞ as x → +0, while fg (x)→ 0 as x → +0.Example 21( x1 )ln x1=lim= lim= 0 for α > 0.x→+∞ x αx→+∞ αx α−1x→+∞ αx αlimExample 22xααx α−1α(α − 1) · · · (α − n + 1)x a−n=···=lim=lim=0x→+∞ a xx→+∞ a x ln ax→+∞a x (ln a)nlimfor a > 1, since for n > α and a > 1 it is obvious thatx α−nax→ 0 if x → +∞.We remark that this entire chain of equalities was hypothetical until we arrivedat an expression whose limit we could find.5.4.5 Constructing the Graph of a FunctionA graphical representation is often used to gain a visualizable description of a function.
As a rule, such a representation is useful in discussing qualitative questionsabout the behavior of the function being studied.For precise computations graphs are used more rarely. In this connection whatis important is not so much a scrupulous reproduction of the function in the formof a graph as the construction of a sketch of the graph of the function that correctlyreflects the main elements of its behavior. In this subsection we shall study somegeneral devices that are encountered in constructing a sketch of the graph of a function.a. Graphs of the Elementary FunctionsWe recall first of all what the graphs of the main elementary functions look like.A complete mastery of these is needed for what follows (Figs. 5.12–5.18).b.
Examples of Sketches of Graphs of Functions (Without Applicationof the Differential Calculus)Let us now consider some examples in which a sketch of the graph of a function canbe easily constructed if we know the graphs and properties of the simplest elementary functions.5.4 Differential Calculus Used to Study FunctionsFig. 5.15Fig. 5.16Fig.
5.17Fig. 5.18Example 23 Let us construct a sketch of the graph of the functionh = logx 2 −3x−2 2.Taking account of the relationy = logx 2 −3x+2 2 =11,=log2 (x 2 − 3x + 2) log2 (x − 1)(x − 2)2512525Differential CalculusFig. 5.19we construct successively the graph of the quadratic trinomial y1 = x 2 − 3x + 2,then y2 = log2 y1 (x), and then y = y21(x) (Fig. 5.19).The shape of this graph could have been “guessed” in a different way: by firstdetermining the domain of definition of the function logx 2 −3x+2 2 = (log2 (x 2 −3x +2))−1 , then finding the behavior of the function under approach to the boundarypoints of the domain of definition and on intervals whose endpoints are the boundarypoints of the domain of definition, and finally drawing a “smooth curve” takingaccount of the behavior thus determined at the ends of the interval.Example 24 The construction of a sketch of the graph of the function y = sin x 2can be seen in Fig.
5.20.We have constructed this graph using certain characteristic points for this function, the points where sin(x 2 ) = −1, sin(x 2 ) = 0, or sin(x 2 ) = 1. Between two adjacent points of this type the function is monotonic. The form of the graph near thepoint x = 0, y = 0 is determined by the fact that sin(x 2 ) ∼ x 2 as x → 0. Moreover,it is useful to note that this function is even.Since we will be speaking only of sketches rather than a precise constructionof the graph of a function, let us agree for the sake of brevity to understand that“constructing the graph of a function” is equivalent to “constructing a sketch of thegraph of the function”.Example 25 Let us construct the graph of the functiony = x + arctan x 3 − 15.4 Differential Calculus Used to Study Functions253Fig.
5.20(Fig. 5.21). As x → −∞ the graph is well approximated by the line y = x −while for x → +∞ it is approximated by y = x + π2 .π2,We now introduce a useful concept.Definition 4 The line c0 + c1 x is called an asymptote of the graph of the functiony = f (x) as x → −∞ (or x → +∞) if f (x) − (c0 + c1 x) = o(1) as x → −∞ (orx → +∞).Thus in the present example the graph has the asymptote y = x − π2 as x → −∞and y = x + π2 as x → +∞.If |f (x)| → ∞ as x → a − 0 (or as x → a + 0) it is clear that the graph of thefunction will move ever closer to the vertical line x = a as x approaches a. We callthis line a vertical asymptote of the graph, in contrast to the asymptotes introducedin Definition 4, which are always oblique.Thus, the graph in Example 23 (see Fig.
5.19) has two vertical asymptotes andone horizontal asymptote (the same asymptote as x → −∞ and as x → +∞).It obviously follows from Definition 4 thatf (x),xc0 = lim f (x) − c1 x .c1 = limx→−∞x→−∞In general, if f (x) − (c0 + c1 x + · · · + cn x n ) = o(1) as x → −∞, then2545Differential CalculusFig. 5.21f (x),xnf (x) − cn x n,cn−1 = limx→−∞x n−1...c0 = lim f (x) − c1 x + · · · + cn x n .cn = limx→−∞x→−∞These relations, written out here for the case x → −∞, are of course also validin the case x → +∞ and can be used to describe the asymptotic behavior of thegraph of a function f (x) using the graph of the corresponding algebraic polynomialc1 + c 1 x + · · · + c n x n .5.4 Differential Calculus Used to Study Functions255Fig.
5.22Example 26 Let (ρ, ϕ) be polar coordinates in the plane and suppose a point ismoving in the plane in such a way thatπt,2πϕ = ϕ(t) = 1 − e−t sin t2ρ = ρ(t) = 1 − e−t cosat time t (t ≥ 0). Draw the trajectory of the point.In order to do this, we first draw the graphs of ρ(t) and ϕ(t) (Figs. 5.22a and5.22b).Then, looking simultaneously at both of the graphs just constructed, we can describe the general form of the trajectory of the point (Fig. 5.22c).c. The Use of Differential Calculus in Constructing the Graph of a FunctionAs we have seen, the graphs of many functions can be drawn in their general featureswithout going beyond the most elementary considerations. However, if we want tomake the sketch more precise, we can use the machinery of differential calculus incases where the derivative of the function being studied is not too complicated.
Weshall illustrate this using examples.Example 27 Construct the graph of the function y = f (x) whenf (x) = |x + 2|e−1/x .The function f (x) is defined for x ∈ R\0. Since e−1/x → 1 as x → ∞, it followsthat!−(x + 2) as x → −∞,−1/x|x + 2|e∼(x + 2)as x → +∞.2565Differential CalculusFig. 5.23Table 5.2Interval]−∞, −2[]−2, 0[Sign of f (x)−+]0, +∞[+Behavior of f (x)+∞ ( 00 ) +∞0 ) +∞Next, it is obvious that |x + 2|e−1/x → +∞ as x → −0, and |x + 2|e−1/x → 0as x → +0.
Finally, it is clear that f (x) ≥ 0 and f (−2) = 0. On the basis of theseobservations, one can already make a first draft of the graph (Fig. 5.23a).Let us now see for certain whether this function is monotonic on the intervals]−∞, −2[, [−2, 0[, and ]0, +∞[, whether it really does have these asymptotics,and whether the convexity of the graph is correctly shown.Since⎧⎨ − x 2 +x+2 e−1/x , if x < −2,x2f (x) =2⎩ x +x+2 e−1/x ,if − 2 < x and x = 0,x2and f (x) = 0, we can form Table 5.2.On the regions of constant sign of the derivative, as we know, the function exhibits the corresponding monotonicity. In the bottom row of the table the symbol+∞ ( 0 denotes a monotonic decrease in the values of the function from +∞ to0, and 0 ) +∞ denotes monotonic increase from 0 to +∞.We observe that f (x) → −4e−1/2 as x → −2 − 0 and f (x) → 4e−1/2 as x →−2 + 0, so that the point (−2, 0) must be a cusp in the graph (a bend of the sametype as in the graph of the function |x|), and not a regular point, as depicted inFig.
5.23a). Next, f (x) → 0 as x → +0, so that the graph should emanate from theorigin tangentially to the x-axis (remember the geometric meaning of f (x)!).We now make the asymptotics of the function as x → −∞ and x → +∞ moreprecise.Since e−1/x = 1 − x −1 + o(x −1 ) as x → ∞, it follows that!−x − 1 + o(1) as x → −∞,−1/x|x + 2|e∼x + 1 + o(1)as x → +∞,5.4 Differential Calculus Used to Study Functions257Table 5.3Interval]−∞, −2[]−2, 0[]0, 2/3[]2/3, +∞[Sign of f (x)−++−Convexity of f (x)UpwardDownwardDownwardUpwardso that in fact the oblique asymptotes of the graph are y = −x − 1 as x → −∞ andy = x + 1 as x → +∞.From these data we can already construct a quite reliable sketch of the graph, butwe shall go further and find the regions of convexity of the graph by computing thesecond derivative:! 2−3x −1/x, if x < −2,− 4 ef (x) = 2−3xx −1/xe,if − 2 < x and x = 0.x4Since f (x) = 0 only at x = 2/3, we have Table 5.3.Since the function is differentiable at x = 2/3 and f (x) changes sign as x passesthrough that point, the point (2/3, f (2/3)) is a point of inflection of the graph.Incidentally, if the derivative f (x) had had a zero, it would have been possibleto judge using the table of values of f (x) whether the corresponding point was anextremum.
In this case, however, f (x) has no zeros, even though the function hasa local minimum at x = −2. It is continuous at that point and f (x) changes fromnegative to positive as x passes through that point. Still, the fact that the functionhas a minimum at x = −2 can be seen just from the description of the variation ofvalues of f (x) on the corresponding intervals, taking into account, of course, therelation f (−2) = 0.We can now draw a more precise sketch of the graph of this function (Fig. 5.23b).We conclude with one more example.Example 28 Let (x, y) be Cartesian coordinates in the plane and suppose a movingpoint has coordinatestt − 2t 3,y=1 − t21 − t2at time t (t ≥ 0). Describe the trajectory of the point.We begin by sketching the graphs of each of the two coordinate functions x =x(t) and y = y(t) (Figs. 5.24a and 5.24b).The second of these graphs is somewhat more interesting than the first, and sowe shall describe how to construct it.We can see the behavior of the function y = y(t) as t → +0, t → 1 − 0, t →1 + 0, and the asymptote y(t) = 2t + o(1) as t → +∞ immediately from the formof the analytic expression for y(t).After computing the derivativex=ẏ(t) =1 − 5t 2 + 2t 4,(1 − t 2 )22585Differential CalculusFig.
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