MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Lecture 2 18.01 Fall 2006 Lecture 2: Limits, Continuity, and Trigonometric Limits More about the “rate of change” interpretation of the derivative y = f(x)yx∆x∆yFigure 1: Graph of a generic function, with Δx and Δy marked on the graph 3. T temperature gradient Δy Δx → dy dx as Δx → 0 Average rate of change → Instantaneous rate of change Examples 1. q = charge dq dt = electrical current 2. s = distance ds dt = speed dT = temperature = dx 1Lecture 2 18.01 Fall 2006 4. Sensitivity of measurements: An example is carried out on Problem Set 1. In GPS, radio signals give us h up to a certain measurement error (See Fig. 2 and Fig. 3). The question is ΔLhow accurately can we measure L. To decide, we find . In other words, these variables are Δh related to each other. We want to find how a change in one variable affects the other variable. LhssatelliteyouFigure 2: The Global Positioning System Problem (GPS) hsLFigure 3: On problem set 1, you will look at this simplified “flat earth” model 2� Lecture 2 18.01 Fall 2006 Limits and Continuity Easy Limits x2 + x 32 + 3 12lim = = = 3 x→3 x + 1 3 + 1 4 With an easy limit, you can get a meaningful answer just by plugging in the limiting value. Remember, lim Δf = lim f(x0 + Δx) − f(x0) x→x0 Δx x→x0 Δx is never an easy limit, because the denominator Δx = 0 is not allowed. (The limit x x0 is computed under the implicit assumption that x =� x0.) → Continuity We say f(x) is continuous at x0 when lim f (x) = f(x0) x x0→Pictures xyFigure 4: Graph of the discontinuous function listed below x + 1 x > 0 f(x) = −x x ≥ 0 3Lecture 2 18.01 Fall 2006 This discontinuous function is seen in Fig. 4. For x > 0, lim f(x) = 1 x 0→but f(0) = 0. (One can also say, f is continuous from the left at 0, not the right.) 1. Removable Discontinuity Figure 5: A removable discontinuity: function is continuous everywhere, except for one point Definition of removable discontinuity Right-hand limit: lim f(x) means lim f (x) for x > x0. +0 x x0→x x→Left-hand limit: lim f (x) means lim f (x) for x < x0. 0x−f(x) = lim f(x) but this is not f(x0), or if f (x0) is undefined, we say the disconti-x x0x→→If lim +00x−→nuity is removable. x x x→For example, sin(xx) is defined for x = 0. We will see later how to evaluate the limit as � x → 0. 4Lecture 2 18.01 Fall 2006 2. Jump Discontinuity x0Figure 6: An example of a jump discontinuity lim for (x < x0) exists, and lim for (x > x0) also exists, but they are NOT equal. +0−x0x x x→ → 3. Infinite Discontinuity yxFigure 7: An example of an infinite discontinuity: 1 x 1 1Right-hand limit: lim = ∞; Left-hand limit: lim x→0+ x x→0− x = −∞ 5Lecture 2 18.01 Fall 2006 4. Other (ugly) discontinuities Figure 8: An example of an ugly discontinuity: a function that oscillates a lot as it approaches the origin This function doesn’t even go to ±∞ — it doesn’t make sense to say it goes to anything. For something like this, we say the limit does not exist. 6Lecture 2 18.01 Fall 2006 Picturing the derivative xyxy’Figure 9: Top: graph of f (x) = 1 and Bottom: graph of f �(x) = − 1 2 x x Notice that the graph of f(x) does NOT look like the graph of f�(x)! (You might also notice that f (x) is an odd function, while f�(x) is an even function. The derivative of an odd function is always even, and vice versa.) 7Lecture 2 18.01 Fall 2006 Pumpkin Drop, Part II This time, someone throws a pumpkin over the tallest building on campus. Figure 10: y = 400 − 16t2 , −5 ≤ t ≤ 5 Figure 11: Top: graph of y(t) = 400 − 16t2 . Bottom: the derivative, y�(t) 8Lecture 2 18.01 Fall 2006 Two Trig Limits Note: In the expressions below, θ is in radians— NOT degrees! lim sin θ = 1; lim 1 − cos θ = 0 θ 0 θ θ 0 θ→ →Here is a geometric proof for the first limit: 1θarclength = θsinθFigure 12: A circle of radius 1 with an arc of angle θ sin θ arclength = θ 1θFigure 13: The sector in Fig. 12 as θ becomes very small Imagine what happens to the picture as θ gets very small (see Fig. 13). As θ 0, we see that sin θ → 1. θ → 9Lecture 2 18.01 Fall 2006 What about the second limit involving cosine? 1cos θ1 - cos θarclength = θθFigure 14: Same picture as Fig. 12 except that the horizontal distance between the edge of the triangle and the perimeter of the circle is marked From Fig. 15 we can see that as θ → 0, the length 1 − cos θ of the short segment gets much smaller than the vertical distance θ along the arc. Hence, 1 − cos θ 0. θ → 1cos θ1 - cos θarclength = θθFigure 15: The sector in Fig. 14 as θ becomes very small 10� � Lecture 2 18.01 Fall 2006 We end this lecture with a theorem that will help us to compute more derivatives next time. Theorem: Differentiable Implies Continuous. If f is differentiable at x0, then f is continuous at x0. f(x) − f (x0)Proof: xlim x0 (f(x) − f (x0)) = xlim x0 x − x0 (x − x0) = f�(x0) · 0 = 0. → →Remember: you can never divide by zero! The first step was to multiply by x − x0 . It looks as 0 x − x0 if this is illegal because when x = x0, we are multiplying by . But when computing the limit as 0 x → x0 we always assume x �= x0. In other words x − x0 �= 0. So the proof is valid.
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