《高等数学》课程电子教案（PPT课件）Chapter 1 Functions and Limits §1.7 Two Remarkable Limits

Chapter1FunctionsLimitsand$ 1.7 Two Remarkable Limits

Chapter 1 Functions and Limits §1.7 Two Remarkable Limits

IntroductionI.SqueezeTheoremsinxlimI.x0Xm.Monotonic SequenceTheoremIVlim(1+=)*=ex-80十S 1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits I. Squeeze Theorem Introduction III. Monotonic Sequence Theorem ) e 1 lim(1+ = → x x x IV 1 sin lim 0 = → x x x II

I.SqueezeTheorem1.SequenceSuppose thatt lim y, = A,lim z, = A,n80, ≤ x, ≤ zn,for n ≥ K (K is a fixed integer)Then lim x, = A.n-→0S 1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits I. Squeeze Theorem 1.Sequence y x z for n K (K is a fixed integer). y A z A n n n n n n n    = = → → , Suppose that lim ,lim , Then x A. n n = → lim

I. Squeeze Theorem2. FunctionSuppose thatlim g(x) = A, lim h(x) = A,x-→ax→ax-→>00X-→01g(x)≤ f(x)≤h(x),as xeU(a,)(x|> M)yThen lim f(x) = A.x-→ax-→00h2gx0aS 1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits 2. Function lim g(x) A,lim h(x) A, x x a x x a = = → → → → Suppose that Then f x A. x x a = → → lim ( ) g(x)  f (x)  h(x),as x U(a, )( x  M). 0 o x y a A g f h I. Squeeze Theorem

Example12"Provethat=0limn!n-→8Proof2"222242.2...202″4Wrong!lim0<limlimn!n-00n-→0n-0nS1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits Example 1 0 ! 2 lim = → n n n Prove that n n n     = 1 2 2 2 2 ! 2 n 2 3 2 2 2 1 2 =    n 4 0   Proof 0, 4 lim 0 = lim = n→ n→ n since 0. ! 2 lim = → n n n therefore n n n n n n 4 lim ! 2 lim 0 lim → → →   Wrong!

Example2Provethatlim cosx = 1x-0xxx0 ≤1-cosx=2sin?2<222Key: Find the expressions of y, and z, (or g(x)and h(x)whose limits are easy to get.S1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits Example 2 lim cos 1 0 = → x x Prove that Key: Find the expressions of whose limits are easy to get. y and z ( or g(x)and h(x)) n n 2 2 2         x 0  2 1 cos 2sin2 x − x = 2 2 x =

Il.RemarkableLimitsinxlimx→0xy=xyy= sinxT元(a)S 1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits II. Remarkable Limit 1 sin lim 0 = → x x x y = x sin x x

Il.RemarkableLimitsinxlimx-→0xysin @(radians)8A3T3T-2T2TTTTTNOTTOSCALES 1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits II. Remarkable Limit 1 sin lim 0 = → x x x

Il.RemarkableLimitDBsinxlim1x-→0xXA0cGeometric proof:S AOB<S<sSecAOBAAOD111sinx<x<tanxtan xsinxx2221tanxxsinx1<<1cosx<sinxsinxcosxxsinxlim cos x = lim 1 = 1Then lim门x-→0x-→0x-→0xS 1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits x o A D C B SAOB  SsecAOB  SAOD Geometric proof: x x tan x 2 1 2 1 sin 2 1   sin x  x  tan x x x x x x cos 1 sin tan sin  1   = 1 sin  cos   x x x lim cos lim1 1 0 0 = = x→ x→ x 1 sin lim 0 = → x x x Then II. Remarkable Limit 1 sin lim 0 = → x x x

Il.RemarkableLimitDBsinxPlim1x→0xA~0Geometric proof:S<SSse POC < SABOCSecAOB11cosx·sinxx.x<COS2221sinx=cosx<xcosx1sinx=1lim cos x = limThen Ilimx-0x-0 cosxx-0xS1.7 Two Remarkable Limits

§1.7 Two Remarkable Limits x o A D C B SsecPOC  SBOC  SsecAOB Geometric proof: x  x  x  x    x 2 2 1 2 1 cos sin 2 1 cos 2 1 x x x x cos sin 1  cos   1 cos 1 lim cos lim 0 0 = = → → x x x x 1 sin lim 0 = → x x x Then P II. Remarkable Limit 1 sin lim 0 = → x x x