第一讲 习题解答习题1-11 计算下列极限① ()1lim 11,0p n n p n →∞⎡⎤⎛⎫+->⎢⎥ ⎪⎝⎭⎢⎥⎣⎦解：原式=()1111110lim lim 110ppp n n n n n n→∞→∞⎛⎫⎛⎫+-+-+ ⎪ ⎪⎝⎭⎝⎭=-()()0110lim 0p p n x x →+-+=-()()01p x x p ='=+= ② ()sin sin limsin x a x a x a →--解：原式=()()()()sin sin sin sin limlimsin x a x a x a x a x ax a x a x a →→---⋅=---=()sin cos x ax a ='= ③1x →，,m n 为自然数 解：原式=11x x n m→='==④()lim 21,0nn a →∞>解：原式()()10ln 21lim ln 211limln 1lim n x n x a e a n nxn ee e →∞→⎛⎫ ⎪⋅- ⎪⎝⎭-→∞====()()()()0ln 21ln 21ln 21lim2ln 20x a a xx a a xx e ee a ---→'-====⑤ lim,0x ax a a x a x a→->- 解：原式=limx a a a x a a a a x x a →-+--lim lim x a a ax a x a a a x a x a x a →→--=---()()x a x a x a a x ==''=-()ln 1a a a =- ⑥ lim ,0xaa xxax a a a a a x →->-解：原式limlim x a x aa x a x x a x a x a x a a a a a x aa x x a a x→→---==⋅---()lim x aa aa a x ax ax a a a a a x ax aa x→----=⋅-- lim xaaaa a x ax a x a a a a a x a x a x a a x →⎛⎫---=-⋅ ⎪ ⎪---⎝⎭lim xaaaa a x a a a a a x a x a a a a a x a x ax a x a x a a x →⎛⎫----=-⋅⋅ ⎪ ⎪----⎝⎭()()()()1ln 1x aa y aa y a x a x a a a x a a ===⎛⎫'''=-⋅⋅ ⎪⎪-⎝⎭ln aa a a =⋅ ⑦ ()()101011sin limsin x tgx x x→+--解：原式=()()101011sin limsin x tgx x xx x→+--⋅()()()()1010101001101sin 1sin 0lim x tgx tg x x x →⎛⎫+-+---=-⎪ ⎪⎝⎭()()()()101011sin x x tgx x ==''=+--20=⑧ ()11lim m k m n i n i kn n -→∞=⎡⎤+-⎢⎥⎢⎥⎣⎦∑，m 为自然数 解：原式()111lim lim 1m m k k m n n i i n i i n n n n n -→∞→∞==⎡⎤⎛⎫⎛⎫+⎛⎫⎢⎥=-=+- ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎢⎥⎝⎭⎝⎭⎣⎦∑∑ ()()()110111lim 12mkk m n i i x i mk k n i i x i n→∞===⎛⎫+- ⎪+'⎝⎭=⋅=⋅+=∑∑2 设()f x 在0x 处二阶可导，计算()()()00022limh f x h f x f x h h→+-+-。

解：()()()()()000002002lim lim 2h h f x h f x f x h f x h f x h h h→→''+-+-+--=()()()()000001lim 2h f x h f x f x h f x h h →''''+---⎡⎤=+⋅⎢⎥-⎣⎦()0f x ''= 3 设()0f x '存在，计算()()000limx x xf x x f x x x →--解：原式()()()()00000000limx x xf x x f x x f x x f x x x →---⎡⎤⎣⎦=-()()()()()00000000limx x f x f x f x x f x x f x x x →-'=-⋅=--4 设()0,a f a '>存在，计算()()1ln ln lim x ax a f x f a -→⎡⎤⎢⎥⎣⎦解：原式()()()()()()ln ln ln ln limlimln ln ln ln x ax aaf a f x f a f x f a x af a x ax ax ae e e→→'---⋅---===习题1-21求下列极限 ①(lim sinx →+∞解：原式()()lim211x x x →+∞=+--(lim sin 20x ξξ→+∞='=⋅=ξ介于1x -，与1x +之间。

② ()40cos sin cos lim sin x x xx→-？？？？解：原式()()40sin sin lim0sin x x x xξξ→--=→ ()32000sin cos 1sin 1limlim lim 366x x x x x x x x x x →→→---+====③ 3320lim sin x xx e e tgx x -→-⋅解：原式()()3333333000lim lim 2lim 22x x x x x x x x e e e e e x x x ξξ--→→→=--'==⋅==-- ④ 2222lim ,0x ax a a x a x a →->- 解：原式2222221lim x aa a x a a a x a x a x ax a →⎛⎫⎛⎫--=-⋅ ⎪ ⎪ ⎪⎪--+⎝⎭⎝⎭()()222211211lim lim ln 22xaa aax x ax a a a ξξξξ→→=='⎛⎫'=-⋅=- ⎪⎝⎭ ⑤ 2lim 1n a a n arctgarctg n n →+∞⎛⎫- ⎪+⎝⎭解：原式()21lim11n a aarctgarctg an n n a a n n n n →+∞-+=⋅+-+()0lim x arctgx a a ξξ→='=⋅= ⑥((()11111lim 1n x x -→-L解：原式1111lim 111x x x x→=---1111lim 111x x x x →=⋅---L11lim !x n →'''=⋅=L⑦limx解：原式lim x x →+∞=lim211x x x →+∞=+-- ⎪⎝⎭11lim 23x ξξ→='=⋅=⑧()2limln 1n n n →∞+解：原式()()11ln 1ln 21lim ln 1n n n n n e e n n ++→∞⎡⎤-⎢⎥⎣⎦=+()()2ln 1ln 1lim ln 1n n n nn n n →∞+⎛⎫- ⎪+⎝⎭=+()()()()21ln 1ln lim 1ln 1n n n n n n n n →∞++-=++()()()()22ln 1ln ln 1lim 1ln 1n n n n n n n n n →∞+-++=++()()()21ln 1ln 1lim 1ln 1n n n n n n n →∞⎛⎫+++ ⎪⎝⎭=++()()()21ln 1lim 1ln 1n n n n n n n →∞++=++()()1ln 1lim1ln 1n n n n n →∞++=⋅++=1 2 设()f x 在a 处可导，()0f a >，计算1lim 1nn f a n f a n →∞⎡⎤⎛⎫+ ⎪⎢⎥⎝⎭⎢⎥=⎛⎫⎢⎥- ⎪⎢⎥⎝⎭⎣⎦解：原式1ln 1lim f a n n f a n n e⎛⎫+ ⎪⎝⎭⎛⎫- ⎪⎝⎭→∞=11ln ln .211lim f a f a n n a a n n n e⎛⎫⎛⎫+-- ⎪ ⎪⎝⎭⎝⎭⎛⎫⎛⎫+-- ⎪ ⎪⎝⎭⎝⎭→∞==()()2f a f ae '习题1-31 求下列极限① ()()011lim ,011x x n x λμ→+-≠+- 解：原式()()011lim 11x x x λλμμ→+-==+-（也可利用对数等价关系求解）②x →解：利用等价代换可得原式0lim x →=()122ln cos cos 2cos lim ln 1x x x nxx→⋅⋅⋅=-+20ln cos ln cos 2ln cos 2limx x x nxx→++⋅⋅⋅+=- 2220cos 1cos 21cos 12lim x x x nx x x x →⎛---⎫⎛⎫=-++⋅⋅⋅+⎪ ⎪⎝⎭⎝⎭ 212222n ⎛⎫⎛⎫=---+⋅⋅⋅+- ⎪ ⎪⎝⎭⎝⎭22212n =++⋅⋅⋅+()()11216n n n =++③ 011lim 1x x x e →⎛⎫-⎪-⎝⎭解：原式()200111lim lim 21x x x x x e x e x x x e →→----===-④ ()112lim 1x x x x x x →∞⎡⎤+-⎢⎥⎣⎦解：原式()()11ln 12ln 211lim lim ln 1ln x x x x x x x e e x x x xx +→+∞→+∞⎡⎤⎡⎤=-=+-⎢⎥⎢⎥⎣⎦⎣⎦()1ln 1lim ln 1ln lim 11x x x x x x x→+∞→+∞⎛⎫+ ⎪⎝⎭=+-==⎡⎤⎣⎦⑤ 3012cos lim 13x x x x→⎡⎤+⎛⎫-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦解：原式()3200ln 2cos ln 312cos 1lim ln lim 36xx x x x x x →→+-+⎛⎫===- ⎪⎝⎭⑥20x x -→解：原式2cos 11x x x e -→---=，而211cos 2x x -:，21x e -→，cos 1x →，()0x →且220112lim 12x xx →=-≠--。

故原式222126112x x x-==- 3计算下列极限 ① 2221cos ln cos limsin x x x x x e ex-→----解：211cos 2x x -:，()()22222222,sin x x e e x x x x x ----=::，且2202lim 21x x x →=-≠--，故原式1=。