Lim x 0 1 cos2x x

• [PDF File]Math 5440 Problem Set 5 – Solutions

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  The solution to the Cauchy problem with u(x,0)=1 for all x is u(x,t)=1 for all x and t >0. The function u(x,t)=1+1 n e n2t sin(nx)is the solution to the Cauchy problem for the initial condition u(x,0)=1+1 n sin(nx). The maximum difference between the initial functions is 1/n which gets smaller and smaller as n grows. The maximum difference

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• [PDF File]Section 1.4 Continuity and One-Sided Limits

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  126. (a) lim1-c°sx _ lira1-cosx 1 + cosx x-*O x2 x-*O x2 1 + cos x = lim 1 -- COS2X x-*°x~(1 + cos x) sinZx 1 = lim x--40 X2 1 + COS X (b) From part (a), 1 - cos x 1~- ~ 1-cosx ~ xl_ z ~ cosx ~ 1-- 1 x2 forx ~ O. xz 2 2 2 (c) cos(0.1) ~ 1 - ~(0.1)2 = 0.995 (d) cos(0.1) ~ 0.9950, which agrees with part (c). 127.

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• [PDF File]Homework3—Due9/20/2010inclass Name: Solutions ...

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  In this case, the given equation has a solution if cosx x2 + 1 = 0, so our function is f(x) = cosx x 2 + 1. The function is a difference of two functions that are continuous

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• [PDF File]2017-06-14 11:46 - Bergen

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  cos2x = cos2 x —sin2 x cosx cosax tan x lim tan ax lim bx a FACTS : EX#I : EX#2: EX#3: sin x lim sin ax lim bx sin 2x lim 5x 2 tan 5x lim 9x 4 sin 3x lim a lim lim EX#4: EX#5 : EX#6: sin x tan x lim 8 sin x cosx lim 3x 7 sin2 xtan2 x lim

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• [PDF File]December 2-4 Review

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  lim x!0 Z 2x x dt t Solution. For any x6= 0, we have Z 2x x dt t = [lnjtj]2x x = lnj2xj lnjxj: Now j2xj= 2jxj, so lnj2xj= ln(2jxj) = ln2 + lnjxj, so Z 2x x dt t = lnj2xj lnjxj= ln2 + lnjxj lnjxj= ln2: Thus the expression we’re taking the limit of doesn’t even depend on x, and lim x!0 Z 2x x dt t = lim x!0 ln2 = ln2: 9. The sine integral ...

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• [PDF File]The Squeeze Theorem: Statement and Example

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  g(x) ≤f(x) ≤h(x) in an interval around c. Then lim x→c g(x) ≤lim x→c f(x) ≤lim x→c h(x) provided those limits exist. When the limits on the upper bound and lower bound are the same, then the function in the middle is \squeezed" into having the same limit. See page 61 in the text for more details. Theorem 1.2 (The Squeeze Theorem).

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• [PDF File]Solution.

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  1 cos2x x2 = lim x!0 1 (1 2sin2 x) x2 = 2 lim x!0 sinx x lim x!0 sinx x (by Thm 3.1(iii) since lim x!0 2 and lim x!0 sinx x exist) = 2 1 1 = 2. Problem 2. (Apostol 4.6.38) Given the formula 1 + x+ x2 + + xn = xn+1 1 x 1 (valid if x 6= 1), determine, by di eren-tiation, formulae for the following sums: (a) 1 + 2x+ 3x2 + + nxn 1, (b) 1 2x+ 2 x2 ...

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• [PDF File]Hw # 11 Solutions - OU Math

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  But lim x! 0 2x=sin2x= 1 and lim x! 0 x=cos2x= 0;therefore, limit on the right side is 1:0 = 0: Therefore, we obtain lim x! 0+ lny= 0: But lim x! 0+ lny= ln( lim x! 0+ y);and hence ln( lim x! 0+ y) = 0 =) lim x! 0+ y= e0 = 1: 1. 7:1# 14 We use integration by parts. Let u = sand dv = 2sds:Then du = dsand v= 1 ln2

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• [PDF File]Sample examinations Calculus II (201-NYB-05) Winter 2009

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  a. lim x→0+ (logx)2 1+x−1 b. lim ... Find the Taylor series of f(x) = cos2x centred at 1 2 π. State the first four non-zeroterms and give the formula for the nth term. Solution outlines 1. sin ` arccos

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• [PDF File]Unit 5 Limits and Derivatives Test Review

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  A. 2.029 B. 3.141 C. 1.820 D. 1.571 12. lim x!0 1 cos2x x2 is A. 2 B. 3 2 C. 4 D. 0 13. A curve is de ned by y = esin2x. Find dy dx. A. esin2x cos2x B. 2esin2x cos2x C. sin2xecos2x D. 4sin4x 14. For what x coordinate(s) does the function de ned by f(x) = 3x5 5x3 8 have a relative maximum? A. 1 only B. 0 and 1 C. 1 only D. 1 and 1

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• [PDF File]2 Analytic functions - MIT Mathematics

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  f0(0) = lim x!0 x x = 1: On the other hand, if we let zgo to 0 along the positive y-axis then f0(0) = lim y!0 i y i y = 1: 1. 2 ANALYTIC FUNCTIONS 2 The limits don’t agree! The problem is that the limit depends on how zapproaches 0. If we came from other directions we’d get other values. There’s nothing to do, but agree that

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• [PDF File]Math 115 Exam #2 Practice Problems

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  lim x→0 sinx2 1−cos2x without using L’Hˆopital’s Rule. Answer: From Problem 2 we know that sinx2 = x2 ...

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• [PDF File]A-Level Mathematics

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  As shown in Figure 1.1, the functions x, sin(x) and tan(x) agree very closely between x =0 and x ˇ0:25 radians. This figure leads us to consider the use of y=x to approximate both y=sin(x)

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• [PDF File]Math 2260 Exam #2 Solutions

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  Then the above limit is equal to lim b!+1 h x 2e2x i b 0 + 1 2 Z b 0 e 2x dx = lim b!+1 x 2e2x 1 4e2x = lim b!+1 2x 1 4e2x b 0 = lim b!+1 2b 1 4e2b 1 4 = lim b!+1 1 4 2b+ 1 4e2b 1 4 lim b!+1 2b+ 1 4e2b Now, both the numerator 2b+ 1 and the denominator 4e2b are going to zero as b!+1, so we can apply L’H^opital’s Rule to see that

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• [PDF File]L™Hopital™s Rule

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  Example 3 Consider lim x!0 1 cos2x 3x2. Let f(x) = 1 cos2x and g(x) = 3x2. Then f(0) = f0(0) = 0 and g(0) = g0(0) = 0. However, lim x!0 f00(x) g00(x) = lim x!0 4cos2x 6 = 2 3: By L™Hopital™s rule, lim x!0 1 cos2x 3x2 = lim x!0 f(x) g(x) = lim x!0 f00(x) g00(x) = 2 3 Exercise 4 Calculate the following limits: (a)lim x!0 sin3 x tan5x (b)lim x ...

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• [PDF File]L’ Hospital Rule

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  (c) f(x) = 2x + sin 2x, g(x) = (2x sin x) esin x ∴ e (xsinxcosx sinx xcosx) 1 cos2x lim g'(x) f'(x) lim x x sinx + + + = →∞ →∞ Since (lim 1 cos2x does not exist, we cannot apply L’hospital rule. x + →∞) 2. (a) () 1 2(1) 2 4 2sin 4 sec 2sin2x sec x lim (LHR) 0/0 form cos2x 1 tanx lim 2 2 2 4 x 4 x = = π π = − − = − π ...

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• [PDF File]Example: lim x-->0 sin 5x/sin2x - MIT OpenCourseWare

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  Example: lim x→0 sin 2x This is similar to an example we saw earlier in the course. Here f(x) = sin5x, g(x) = sin2x, and a = 0. Since f(a) = g(a) = sin 0 = 0, we can apply l’Hˆopital’s rule and find this limit: sin 5x 5 cos 5x lim x→0 sin 2x = lim x→0 2 cos 2x (l’Hop) = lim 5 cos(5 · 0) x→0 2 cos(2 · 0) 5 = . 2 1

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• [PDF File]10 Fourier Series - UCL

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  0 = 1 π Z π −π f(x)dx a n = 1 π Z π −π f(x)cos(nx)dx b n = 1 π Z π −π f(x)sin(nx)dx then the trigonometric series 1 2 a 0 + X∞ n=1 (a ncos(nx)+b nsin(nx)) is called the Fourier series associated to the function f(x). Remark. Notice that we are not saying f(x) is equal to its Fourier Series. Later we will discuss conditions ...

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• [PDF File]Extra limits with trigonometric functions

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  1 cos2x 3tan2 x = 2=3. 10. Show lim x!0 cos2 x 1 sinx = 1. 11. Show lim x!0 tan2x x 3x sinx = 1=2. 12. Show lim x!a sinx sina x a = cos(a). 13. Show lim x!0 sin5x sin3x sinx = 2. 14. Show lim x!0 tan3x 2x 3x sin2 x = 2. 15. Show lim x!0 x2 tan2x tanx = 1=3. 16. Show lim x!ˇ=4 1 tanx x ˇ=4 = 2. 17. Show lim x!0 tan(x=2) 3x = 2. 18. Show lim x ...

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• [PDF File]I. The Limit Laws

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  • If r > 0 is a rational number then 0 1 lim = x →∞ xr • If r > 0 is a rational number such that xr is defined for all x then 0 1 lim = x →−∞ xr Definition in Words Precise Mathematical Definition Large POSITIVE numbers Let f be a function defined on some interval (a, ∞). Then f x L x = →∞ lim ( ) means that the values of f ...

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