alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Left right also value fx f0 1 0 something undefined

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Unformatted text preview: 1) . (1) -2 δt→ 0 t → t1− We know from the projectile equation for t0 = 0 we have t1 = 2u0.sinθ0 / g . Substituting t0 = 0 and t1 = 2u0.sinθ0 / g in equation (1) above, we get : L imit y(t) = L imit y (t) = Limit 0 →− →− δt→ t t1 t t1 0 y0(t) = 0 - - - (2) . Approaching t1 from the right y(t) is : y1(t) = u1.sinθ1.(t − t1) − 1 g (t − t1)2. right -2 Near t1 and to the right of t1 : t = t1 + δt . right So, y1(t) = u1.sinθ1.(t1 + δt − t1)− 1 g (t1 + δt − t1)2 = u1.sinθ1.( δt ) − 1 g ( δt)2 --2 2 L imit y(t) = Limit y (t) = Limit { u .sinθ .( δt ) − 1 g ( δt)2 } = 0 . -δt → 0 1 1 t → t1+ t → t1+ 1 2 Hence, L imit − y(t) = 0 = L imit + y(t) t → t1 t → t1 Value y(t) = 0 t= t 1 Since, L imit t → t1− we say : y(t) = L imit + y(t) = tValue y(t) = 0 t→t =t 1 1 y(t) is continuous at t = t1. continuous 48 Example 5 : is f(x) = + √x continuous at x = 0 ? continuous y y = +√ x (0, 0) x Near 0 and to the left of 0 : left L imit x → 0− x = 0 − δx Limit √ (0 − δx) = Limit √ − δx δx → 0 f(x) = δx → 0 A negative real number under the radical sign is a complex number. We are dealing with real numbers and real valued functions. Here √ − δx is not real valued and so it is not defined. f(x) = + √x is not defined for x < 0. not L imit x → 0+ f(x) = Limit √ (0 +δx) = Limit √+ δx = 0 δx → 0 δx → 0 Since, L imit − f(x) ≠ L imit + f(x) , we say L imit f(x) does not exist. x→0 x→0 x→0 Value f(x) = f(0) = √0 = 0 x=0 Since L imit → x 0 f(x) does not exist we say : f(x) is discontinuous at x = 0. discontinuous We may remove this discontinuity by redefining f(x) to be : f(x) = { 49 0 for x < 0 √x for x ≥ 0 Example 6: Is the function f(x) = 1 x continuous at x = 0 ? / continuous Limit from the LEFT L imit { f(x) } = L imit { 1 } = − 1 = −∞ , something undefined. /x /0 x → 0− x → 0− Limit { f(x) } = Limit {f(0 − δ x)} = Limit { 1 / (0 − δ x) } = − 1 0 = −∞ /...
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