( 1 ) lim x → 0 l n ( 1 + x ) x ； ( 2 ) lim x → 0 e x − e − x s i n x ； ( 3 ) lim x → 0 t a n x − x x − s i n x ； ( 4 ) lim x → π s i n 3 x t a n 5 x ； ( 5 ) lim x → π 2 l n s i n x ( π − 2 x ) 2 ； ( 6 ) lim x → a x m − a m x n − a n ( a ≠ 0 ) ； ( 7 ) lim x → 0 + l n t a n 7 x l n t a n 2 x ； ( 8 ) lim x → π 2 t a n x t a n 3 x ； ( 9 ) lim x → + ∞ l n ( 1 + 1 x ) a r c c o t x ； ( 10 ) lim x → 0 l n ( 1 + x 2 ) s e c x − c o s x ； ( 11 ) lim x → 0 x c o t 2 x ； ( 12 ) lim x → 0 x 2 e 1 x 2 ； ( 13 ) lim x → 1 ( 2 x 2 − 1 − 1 x − 1 ) ； ( 14 ) lim x → ∞ ( 1 + a x ) x ； ( 15 ) lim x → 0 + x s i n x ； ( 16 ) lim x → 0 + ( 1 x ) t a n x (1) limx→0ln(1+x)x； (2) limx→0ex−e−xsin x； (3) limx→0tan x−xx−sin x； (4) limx→πsin 3xtan 5x； (5) limx→π2ln sin x(π−2x)2； (6) limx→axm−amxn−an (a≠0)； (7) limx→0+ln tan 7xln tan 2x； (8) limx→π2tan xtan 3x； (9) limx→+∞ln(1+1x)arccot x； (10) limx→0ln(1+x2)sec x−cos x； (11) limx→0xcot 2x； (12) limx→0x2e1x2； (13) limx→1(2x2−1−1x−1)； (14) limx→∞(1+ax)x； (15) limx→0+xsin x； (16) limx→0+(1x)tan x (1) x→0limxln(1+x)； (2) x→0limsin xex−e−x； (3) x→0limx−sin xtan x−x； (4) x→πlimtan 5xsin 3x； (5) x→2πlim(π−2x)2ln sin x； (6) x→alimxn−anxm−am (a=0)； (7) x→0+limln tan 2xln tan 7x； (8) x→2πlimtan 3xtan x； (9) x→+∞limarccot xln(1+x1)； (10) x→0limsec x−cos xln(1+x2)； (11) x→0limxcot 2x； (12) x→0limx2ex21； (13) x→1lim(x2−12−x−11)； (14) x→∞lim(1+xa)x； (15) x→0+limxsin x； (16) x→0+lim(x1)tan x
( 1 ) lim x → 0 l n ( 1 + x ) x = lim x → 0 1 1 + x = 1 ( 2 ) lim x → 0 e x − e − x s i n x = lim x → 0 e x + e − x c o s x = 2 ( 3 ) lim x → 0 t a n x − x x − s i n x = lim x → 0 s e c 2 x − 1 1 − c o s x = lim x → 0 1 + c o s x c o s 2 x = 2 ( 4 ) lim x → π s i n 3 x t a n 5 x = lim x → π 3 c o s 3 x 5 s e c 2 5 x = lim x → π 3 5 c o s 3 x ⋅ c o s 2 5 x = − 3 5 ( 5 ) lim x → π 2 l n s i n x ( π − 2 x ) 2 = lim x → π 2 c o s x s i n x 8 x − 4 π = lim x → π 2 c o t x 8 x − 4 π = lim x → π 2 − c s c 2 x 8 = − 1 8 ( 6 ) lim x → a x m − a m x n − a n = lim x → a m x m − 1 n x n − 1 = lim x → a m n x m − n = m n a m − n ( 7 ) lim x → 0 + l n t a n 7 x l n t a n 2 x = lim x → 0 + 7 s e c 2 7 x t a n 7 x 2 s e c 2 2 x t a n 2 x = lim x → 0 + 7 s i n 4 x 2 s i n 14 x = lim x → 0 + 28 c o s 4 x 28 c o s 14 x = 1 ( 8 ) lim x → π 2 t a n x t a n 3 x = lim x → π 2 s e c 2 x 3 s e c 2 3 x = lim x → π 2 c o s 2 3 x 3 c o s 2 x = lim x → π 2 − 6 c o s 3 x s i n 3 x − 6 c o s x s i n x = lim x → π 2 s i n 6 x s i n 2 x = lim x → π 2 6 c o s 6 x 2 c o s 2 x = 3 ( 9 ) lim x → + ∞ l n ( 1 + 1 x ) a r c c o t x = lim x → + ∞ − 1 x + x 2 − 1 1 + x 2 = lim x → + ∞ 1 + 1 x 2 1 + 1 x = 1 ( 10 ) lim x → 0 l n ( 1 + x 2 ) s e c x − c o s x = lim x → 0 l n ( 1 + x 2 ) s i n x t a n x = lim x → 0 2 x 1 + x 2 c o s x t a n x + s i n x s e c 2 x = lim x → 0 2 x 1 + x 2 s i n x + t a n x s e c x = lim x → 0 2 ( 1 − x 2 ) ( 1 + x 2 ) 2 c o s x + s e c 3 x + s e c x t a n 2 x = 1 ( 11 ) lim x → 0 x c o t 2 x = lim x → 0 x c o s 2 x s i n 2 x = lim x → 0 c o s 2 x − 2 x s i n 2 x 2 c o s 2 x = lim x → 0 ( 1 2 − x t a n 2 x ) = 1 2 ( 12 ) lim x → 0 x 2 e 1 x 2 = lim x → 0 e 1 x 2 1 x 2 = lim x → 0 e 1 x 2 = + ∞ ( 13 ) lim x → 1 ( 2 x 2 − 1 − 1 x − 1 ) = lim x → 1 2 − ( x + 1 ) ( x + 1 ) ( x − 1 ) = lim x → 1 − 1 x + 1 = − 1 2 ( 14 ) 令 1 t = a x ，则 x = t a ，当 x → ∞ 时， t → ∞ ， lim x → ∞ ( 1 + a x ) x = lim t → ∞ ( 1 + 1 t ) t a = [ lim t → ∞ ( 1 + 1 t ) t ] a = e a ( 15 ) lim x → 0 + x s i n x = e lim x → 0 + s i n x l n x ，而 lim x → 0 + s i n x l n x = lim x → 0 + s i n x x ⋅ l n x 1 x = lim x → 0 + 1 x − 1 x 2 = lim x → 0 + ( − x ) = 0 ， 所以 lim x → 0 + x s i n x = e 0 = 1 ( 16 ) lim x → 0 + ( 1 x ) t a n x = e lim x → 0 + t a n x l n 1 x ，而 lim x → 0 + t a n x l n 1 x = lim x → 0 + t a n x x ⋅ − l n x 1 x = lim x → 0 + − 1 x − 1 x 2 = 0 ， 所以 lim x → 0 + ( 1 x ) t a n x = e 0 = 1 (1) limx→0ln(1+x)x=limx→011+x=1 (2) limx→0ex−e−xsin x=limx→0ex+e−xcos x=2 (3) limx→0tan x−xx−sin x=limx→0sec2 x−11−cos x=limx→01+cos xcos2 x=2 (4) limx→πsin 3xtan 5x=limx→π3cos 3x5sec2 5x=limx→π35cos 3x⋅cos2 5x=−35 (5) limx→π2ln sin x(π−2x)2=limx→π2cos xsin x8x−4π=limx→π2cot x8x−4π=limx→π2−csc2 x8=−18 (6) limx→axm−amxn−an=limx→amxm−1nxn−1=limx→amnxm−n=mnam−n (7) limx→0+ln tan 7xln tan 2x=limx→0+7sec2 7xtan 7x2sec2 2xtan 2x=limx→0+7sin 4x2sin 14x=limx→0+28cos 4x28cos 14x=1 (8) limx→π2tan xtan 3x=limx→π2sec2 x3sec2 3x=limx→π2cos2 3x3cos2 x=limx→π2−6cos 3xsin 3x−6cos xsin x=limx→π2sin 6xsin 2x=limx→π26cos 6x2cos 2x=3 (9) limx→+∞ln(1+1x)arccot x=limx→+∞−1x+x2−11+x2=limx→+∞1+1x21+1x=1 (10) limx→0ln(1+x2)sec x−cos x=limx→0ln(1+x2)sin xtan x=limx→02x1+x2cos xtan x+sin xsec2 x=limx→02x1+x2sin x+tan xsec x= limx→02(1−x2)(1+x2)2cos x+sec3 x+sec xtan2 x=1 (11) limx→0xcot 2x=limx→0xcos 2xsin 2x=limx→0cos 2x−2xsin 2x2cos 2x=limx→0(12−xtan 2x)=12 (12) limx→0x2e1x2=limx→0e1x21x2=limx→0e1x2=+∞ (13) limx→1(2x2−1−1x−1)=limx→12−(x+1)(x+1)(x−1)=limx→1−1x+1=−12 (14) 令1t=ax，则x=ta，当x→∞时，t→∞， limx→∞(1+ax)x=limt→∞(1+1t)ta=[limt→∞(1+1t)t]a=ea (15) limx→0+xsin x=elimx→0+sin xln x，而limx→0+sin xln x=limx→0+sin xx⋅ln x1x=limx→0+1x−1x2=limx→0+(−x)=0， 所以limx→0+xsin x=e0=1 (16) limx→0+(1x)tan x=elimx→0+tan xln 1x，而limx→0+tan xln 1x=limx→0+tan xx⋅−ln x1x=limx→0+−1x−1x2=0， 所以limx→0+(1x)tan x=e0=1 (1) x→0limxln(1+x)=x→0lim1+x1=1 (2) x→0limsin xex−e−x=x→0limcos xex+e−x=2 (3) x→0limx−sin xtan x−x=x→0lim1−cos xsec2 x−1=x→0limcos2 x1+cos x=2 (4) x→πlimtan 5xsin 3x=x→πlim5sec2 5x3cos 3x=x→πlim53cos 3x⋅cos2 5x=−53 (5) x→2πlim(π−2x)2ln sin x=x→2πlim8x−4πsin xcos x=x→2πlim8x−4πcot x=x→2πlim8−csc2 x=−81 (6) x→alimxn−anxm−am=x→alimnxn−1mxm−1=x→alimnmxm−n=nmam−n (7) x→0+limln tan 2xln tan 7x=x→0+limtan 2x2sec2 2xtan 7x7sec2 7x=x→0+lim2sin 14x7sin 4x=x→0+lim28cos 14x28cos 4x=1 (8) x→2πlimtan 3xtan x=x→2πlim3sec2 3xsec2 x=x→2πlim3cos2 xcos2 3x=x→2πlim−6cos xsin x−6cos 3xsin 3x=x→2πlimsin 2xsin 6x=x→2πlim2cos 2x6cos 6x=3 (9) x→+∞limarccot xln(1+x1)=x→+∞lim−1+x21−x+x21=x→+∞lim1+x11+x21=1 (10) x→0limsec x−cos xln(1+x2)=x→0limsin xtan xln(1+x2)=x→0limcos xtan x+sin xsec2 x1+x22x=x→0limsin x+tan xsec x1+x22x= x→0limcos x+sec3 x+sec xtan2 x(1+x2)22(1−x2)=1 (11) x→0limxcot 2x=x→0limsin 2xxcos 2x=x→0lim2cos 2xcos 2x−2xsin 2x=x→0lim(21−xtan 2x)=21 (12) x→0limx2ex21=x→0limx21ex21=x→0limex21=+∞ (13) x→1lim(x2−12−x−11)=x→1lim(x+1)(x−1)2−(x+1)=x→1limx+1−1=−21 (14) 令t1=xa，则x=ta，当x→∞时，t→∞， x→∞lim(1+xa)x=t→∞lim(1+t1)ta=[t→∞lim(1+t1)t]a=ea (15) x→0+limxsin x=ex→0+limsin xln x，而x→0+limsin xln x=x→0+limxsin x⋅x1ln x=x→0+lim−x21x1=x→0+lim(−x)=0， 所以x→0+limxsin x=e0=1 (16) x→0+lim(x1)tan x=ex→0+limtan xln x1，而x→0+limtan xln x1=x→0+limxtan x⋅x1−ln x=x→0+lim−x21−x1=0， 所以x→0+lim(x1)tan x=e0=1
lim x → ∞ x + s i n x x = lim x → ∞ ( 1 + s i n x x ) = 1 limx→∞x+sin xx=limx→∞(1+sin xx)=1 x→∞limxx+sin x=x→∞lim(1+xsin x)=1
lim x → 0 x 2 s i n 1 x s i n x = lim x → 0 x s i n 1 x s i n x x = 0 limx→0x2sin 1xsin x=limx→0xsin 1xsin xx=0 x→0limsin xx2sin x1=x→0limxsin xxsin x1=0
lim x → 0 + f ( x ) = lim x → 0 + [ ( 1 + x ) 1 x e ] 1 x = e lim x → 0 + 1 x l n [ ( 1 + x ) 1 x e ] ， 而 lim x → 0 + 1 x l n [ 1 x l n ( 1 + x ) − 1 ] = lim x → 0 + l n ( 1 + x ) − x x 2 = lim x → 0 + 1 1 + x − 1 2 x = lim x → 0 + − 1 2 ( 1 + x ) = − 1 2 ，所以 lim x → 0 + f ( x ) = e − 1 2 ， 因 lim x → 0 − f ( x ) = lim x → 0 − e − 1 2 = e − 1 2 ， f ( 0 ) = e − 1 2 ，所以 lim x → 0 + f ( x ) = lim x → 0 − f ( x ) = f ( 0 ) ，函数 f ( x ) 在 x = 0 处连续。 limx→0+f(x)=limx→0+[(1+x)1xe]1x=elimx→0+1xln[(1+x)1xe]， 而limx→0+1xln[1xln(1+x)−1]=limx→0+ln(1+x)−xx2=limx→0+11+x−12x=limx→0+−12(1+x)=−12，所以limx→0+f(x)=e−12， 因limx→0−f(x)=limx→0−e−12=e−12，f(0)=e−12，所以limx→0+f(x)=limx→0−f(x)=f(0)，函数f(x)在x=0处连续。 x→0+limf(x)=x→0+lim[e(1+x)x1]x1=ex→0+limx1ln[e(1+x)x1]， 而x→0+limx1ln[x1ln(1+x)−1]=x→0+limx2ln(1+x)−x=x→0+lim2x1+x1−1=x→0+lim−2(1+x)1=−21，所以x→0+limf(x)=e−21， 因x→0−limf(x)=x→0−lime−21=e−21，f(0)=e−21，所以x→0+limf(x)=x→0−limf(x)=f(0)，函数f(x)在x=0处连续。