MATH 1225 Lecture 7: Sec 2.3 (2) (2)

A useful remark: Since the derivative function is defined through limits, it is easy to show the following result: If f(s) and g(x) are equal on an open interval (a, b), and f?(c) exists for some c E (a, b), then g?(c) also exists and g(c) = f (c). this allows one to calculate the derivative of functions Defined piecewise, except at the x values where the definition changes from one formula to another. For example: if then h (x) = 4x on (- infinity , 5) (because it agrees with (2x^2 + 4)? there) and h?(x) = 2e^2x on (5, infinity) (because it agrees with (e^2x) there). Whether or not h?(5) exists requires more work and we will see how to answer this in the following two questions. (1) Consider the function (a) Show that f(s) is continuous at x = 0 by calculating the left and right hand limits of f(s) as it approaches 0 and showing they are both equal to f(0). (b) Show that f(s) is not differentiable at 0 by calculating and showing the two sided limit does not exist (note: calculate these limits explicitly, do not take shortcuts using derivative formulas). These one - sided limits are called the left and right - sided derivatives of f(x) at 0. By definition, f?(0) exists if and only if the two one - sided derivatives both exists and are equal (note: there is nothing special about 0; one sided derivatives are defined similarity for any s value). (c) Using the Power Rule Theorem from class and the remark preceding this question, calculate f(x) (note that by (b), 0 should not be in the domain of .f?(x)). Sketch a graph of f(s) and f(x) on the same XY - axis (you do not need to be precise, just sketch time shape).

1. 0/6 points LarCalc11 13.1.010. [3864 VIA Evaluate the function at the given values of the independent variables. Simplify the results. fx, y) -4 -x2 - 4y2 (a) ro, 0) for (c) f2, 3) (e) x, o Cal 2. 0/2 points LarCalc11 13.1.019.MI. [386 Evaluate the function at the given values of the independent variables. Simplify the results. x, y)- 6x y2 (a) fix + ax, y)-f(x, y) Ay

3. 1 points LarCalc11 13.2 Find the indicated limit by using the limits limfx, y) 2 and im g(x, y)--4 lim 4. O/1 points LarCalci1 13 2 Find the indicated limit by using the limits lim x, y) 3 and lim (x, y) g(x, y)--5 (a, b) lim x, Y) (x, y) (a, b) g(x, y) 5. y1 points LarCalc11 13.3 Explain whether or not the Quotient Rule should be used to find the partial derivative. Do not differentiate. ay x2 51 O Yes, the function is a quotient. O No, y only occurs in the numerator. 6. 0/2 points LarCalc11 13.3 Find both first partial derivatives oz 7. 0/2 points LarCalc11 13.3 Find both first partial derivatives. hx, y) ex+y) h(x, y)

8. /2 points LarCalc11 13.3.031 Find both first partial derivatives. az dx az ay 9. /1 points LarCalc11 13.4.003. Find the total differential. z 3xSy10 dz 10. 0/2 points LarCalc11 13.5.012 Consider the following. (a) Find dw/dt using the appropriate Chain Rule. dw dt (b) Find dw/dt by converting w to a function of t before differentiating. dw dt 11. 0/1 points LarCalc11 13.6.008. Find the directional derivative of the function at P in the direction of v. rx, y) = x3-y3, Rio, 9), v-V2 (1+j)

12. 0/2 points LarCalc11 13.7.017 (a) Find an equation of the tangent plane to the surface at the given point. x+y+z=9, (2, 5, 2) (b) Find a set of symmetric equations for the normal line to the surface at the given point. ox -2y -5-z-2 13. /3 points LarCalc11 13.8.025 Use a computer algebra system to graph the surface and locate any relative extrema and saddle points. (If an an not exist, enter DNE.) -8x x2 y21 relative minimum (x, y, z) - relative maximum (x, y, z) saddle point 14. 0/1 points LarCalci1 13.10.027 Use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. Surface Point Plane: x yz 1 (7,1,1)

42 Metric Spaces Proof. The proof is somewhat lengthy but straightforward. We subdivide it into four steps (a) to (d). We construct (a) X=(X, d) (b) an isometry T of X onto W, where W Then we prove: (c) completeness of X. (d) uniqueness of X, except for isometries. Roughly speaking, our task will be the assignment of suitable limits to Cauchy sequences in X that do not converge. However, we should not introduce "too many" limits, but take into account that certain se- quences "may want to converge with the same limit" since the terms o those sequences "ultimately come arbitrarily close to each other intuitive ide " This a can be expressed mathematically in terms of a suitable equivalence relation [see (1), below). This is not artificial but is suggested by the process of completion of the rational line mentioned at the beginning of the section. The details of the proof are as follows. X-(X, d). (%) and (x) be (a) Construction of Let Cauchy sequences in Х. Define (4) to be equivalente, to (x), written Let X be the set of all equivalence classes,of Cauchy sequences thus obtained wewrite.(%)eito mean that (x.) is a member of i (a representative of the class , we now set where (t)e i and (y)e y. We show that this limit exists. We have hence we obtain and a similar inequality with m and n interchanged. Together, 1e review of the concept of equivalence, see Al 4 in Appendix 1

1.6 Completion of Metric Spaces 43 Since (%) and (%) are Cauchy, we can make the right side as small as we please. This implies that the limit in (2) exists because Ris complete. We must also show that the limit in (2) is independent of the particular choice of representatives. In fact, if ( )-(x) and (%)-(y"), then by (1), as n-a, which implies the assertion We prove that d in (2) is a metric on X. Obviously, d satisfies (MI) in Sec. 1.1 as well as d(i,i)s0 and (M3). Furthermore, gives (M2), and (M4) for d follows from TU ) by letting n-00, with (b) Construction of an isometry T: X-Wc each bex we associate the class bex which contains the constant Cauchy sequence (b, b,). This defines a mapping T X_W onto the subspace W= TX) ,. The mapping T is given by b-→ b= T, where (b, b,.Jeb. We see that T is an isometry since (2) becomes simply メ here é is the class of (ya) where y- c for all n. Any isometry is injective, and T: X-→ W is surjective since T(X)=w. Hence W and X are isometric; cf. Def. 1.6-1(b). We show that W is dense in X. We consider any iex. Let ()e i. For every e>0 there is an N such that in> N)

Let (쯔스-k h. Then This shows that every s-neighborhood of the arbitrary i e X contains an element of W. Hence W is dense in X. (c) Completeness of X. Let () be any Cauchy sequence in X. Since W is dense in X, for every there is a &ne W such that Pl Hence by the triangle inequality and this is less than any given ε>0 for sufficiently large m and n because (辶) is Cauchy. Hence (2-) is Cauchy. Since T: X-w is isometric and i W, the sequence (z), where zm-Tim, is Cauchy in X. Let te X be the class to which (.) belongs. We show that i is the limit of () By (4) rt Since (m)e i (see right before) and i e W, so that (2 the inequality (5) becomes ei nm to and the right side is smaller than any given e>0 for sufficiently large n. Hence the arbitrary Cauchy sequence (L) in X has the limit ieX, and X is complete.

1.6 Completion of Metric Spaces 45 っn+ワ Fig. 11. Notations in part (d) of the proof of Theorem 1.6-2 (d) Uniqueness of X except for isometries. If (X, d) is another complete metric space with a subspace W dense in X and isometric with X, then for any i, fe X we have sequences ( ), (SJ in W such thatn and -y; hence follows from [the inequality being similar to (3)1. Since W is isometric with We X and W-X, the distances on X and X must be the same. Hence X and X are isometric. We shall see in the next two chapters (in particular in 2.3-2, 3.1- and 3.2-3) that this theorem has basic applications to individual incomplete spaces as well as to whole classes of such spaces.