( 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 \begin{aligned} &\ \ (1)\ \ \lim_{x \rightarrow 0}\frac{ln(1+x)}{x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ \lim_{x \rightarrow 0}\frac{e^x-e^{-x}}{sin\ x}；\\\\ &\ \ (3)\ \ \lim_{x \rightarrow 0}\frac{tan\ x-x}{x-sin\ x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ \lim_{x \rightarrow \pi}\frac{sin\ 3x}{tan\ 5x}；\\\\ &\ \ (5)\ \ \lim_{x \rightarrow \frac{\pi}{2}}\frac{ln\ sin\ x}{(\pi-2x)^2}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ \lim_{x \rightarrow a}\frac{x^m-a^m}{x^n-a^n}\ (a \neq 0)；\\\\ &\ \ (7)\ \ \lim_{x \rightarrow 0^+}\frac{ln\ tan\ 7x}{ln\ tan\ 2x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ \lim_{x \rightarrow \frac{\pi}{2}}\frac{tan\ x}{tan\ 3x}；\\\\ &\ \ (9)\ \ \lim_{x \rightarrow +\infty}\frac{ln\left(1+\frac{1}{x}\right)}{arccot\ x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (10)\ \ \lim_{x \rightarrow 0}\frac{ln(1+x^2)}{sec\ x-cos\ x}；\\\\ &\ \ (11)\ \ \lim_{x \rightarrow 0}xcot\ 2x；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (12)\ \ \lim_{x \rightarrow 0}x^2e^{\frac{1}{x^2}}；\\\\ &\ \ (13)\ \ \lim_{x \rightarrow 1}\left(\frac{2}{x^2-1}-\frac{1}{x-1}\right)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ (14)\ \ \lim_{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^x；\\\\ &\ \ (15)\ \ \lim_{x \rightarrow 0^+}x^{sin\ x}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (16)\ \ \lim_{x \rightarrow 0^+}\left(\frac{1}{x}\right)^{tan\ x} & \end{aligned} ​  (1)  x→0lim​xln(1+x)​；                              (2)  x→0lim​sin xex−e−x​；  (3)  x→0lim​x−sin xtan x−x​；                              (4)  x→πlim​tan 5xsin 3x​；  (5)  x→2π​lim​(π−2x)2ln sin x​；                             (6)  x→alim​xn−anxm−am​ (a=0)；  (7)  x→0+lim​ln tan 2xln tan 7x​；                             (8)  x→2π​lim​tan 3xtan x​；  (9)  x→+∞lim​arccot xln(1+x1​)​；                          (10)  x→0lim​sec x−cos xln(1+x2)​；  (11)  x→0lim​xcot 2x；                                  (12)  x→0lim​x2ex21​；  (13)  x→1lim​(x2−12​−x−11​)；              (14)  x→∞lim​(1+xa​)x；  (15)  x→0+lim​xsin 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 \begin{aligned} &\ \ (1)\ \lim_{x \rightarrow 0}\frac{ln(1+x)}{x}=\lim_{x \rightarrow 0}\frac{1}{1+x}=1\\\\ &\ \ (2)\ \lim_{x \rightarrow 0}\frac{e^x-e^{-x}}{sin\ x}=\lim_{x \rightarrow 0}\frac{e^x+e^{-x}}{cos\ x}=2\\\\ &\ \ (3)\ \lim_{x \rightarrow 0}\frac{tan\ x-x}{x-sin\ x}=\lim_{x \rightarrow 0}\frac{sec^2\ x-1}{1-cos\ x}=\lim_{x \rightarrow 0}\frac{1+cos\ x}{cos^2\ x}=2\\\\ &\ \ (4)\ \lim_{x \rightarrow \pi}\frac{sin\ 3x}{tan\ 5x}=\lim_{x \rightarrow \pi}\frac{3cos\ 3x}{5sec^2\ 5x}=\lim_{x \rightarrow \pi}\frac{3}{5}cos\ 3x \cdot cos^2\ 5x=-\frac{3}{5}\\\\ &\ \ (5)\ \lim_{x \rightarrow \frac{\pi}{2}}\frac{ln\ sin\ x}{(\pi-2x)^2}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{\frac{cos\ x}{sin\ x}}{8x-4\pi}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{cot\ x}{8x-4\pi}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{-csc^2\ x}{8}=-\frac{1}{8}\\\\ &\ \ (6)\ \lim_{x \rightarrow a}\frac{x^m-a^m}{x^n-a^n}=\lim_{x \rightarrow a}\frac{mx^{m-1}}{nx^{n-1}}=\lim_{x \rightarrow a}\frac{m}{n}x^{m-n}=\frac{m}{n}a^{m-n}\\\\ &\ \ (7)\ \lim_{x \rightarrow 0^+}\frac{ln\ tan\ 7x}{ln\ tan\ 2x}=\lim_{x \rightarrow 0^+}\frac{\frac{7sec^2\ 7x}{tan\ 7x}}{\frac{2sec^2\ 2x}{tan\ 2x}}=\lim_{x \rightarrow 0^+}\frac{7sin\ 4x}{2sin\ 14x}=\lim_{x \rightarrow 0^+}\frac{28cos\ 4x}{28cos\ 14x}=1\\\\ &\ \ (8)\ \lim_{x \rightarrow \frac{\pi}{2}}\frac{tan\ x}{tan\ 3x}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{sec^2\ x}{3sec^2\ 3x}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{cos^2\ 3x}{3cos^2\ x}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{-6cos\ 3xsin\ 3x}{-6cos\ xsin\ x}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{sin\ 6x}{sin\ 2x}=\lim_{x \rightarrow \frac{\pi}{2}}\frac{6cos\ 6x}{2cos\ 2x}=3\\\\ &\ \ (9)\ \lim_{x \rightarrow +\infty}\frac{ln\left(1+\frac{1}{x}\right)}{arccot\ x}=\lim_{x \rightarrow +\infty}\frac{-\frac{1}{x+x^2}}{-\frac{1}{1+x^2}}=\lim_{x \rightarrow +\infty}\frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}=1\\\\ &\ \ (10)\ \lim_{x \rightarrow 0}\frac{ln(1+x^2)}{sec\ x-cos\ x}=\lim_{x \rightarrow 0}\frac{ln(1+x^2)}{sin\ x tan\ x}=\lim_{x \rightarrow 0}\frac{\frac{2x}{1+x^2}}{cos\ x tan\ x+sin\ x sec^2\ x}=\lim_{x \rightarrow 0}\frac{\frac{2x}{1+x^2}}{sin\ x+tan\ x sec\ x}=\\\\ &\ \ \ \ \ \ \ \ \ \ \lim_{x \rightarrow 0}\frac{\frac{2(1-x^2)}{(1+x^2)^2}}{cos\ x+sec^3\ x+sec\ xtan^2\ x}=1\\\\ &\ \ (11)\ \lim_{x \rightarrow 0}xcot\ 2x=\lim_{x \rightarrow 0}\frac{xcos\ 2x}{sin\ 2x}=\lim_{x \rightarrow 0}\frac{cos\ 2x-2xsin\ 2x}{2cos\ 2x}=\lim_{x \rightarrow 0}\left(\frac{1}{2}-xtan\ 2x\right)=\frac{1}{2}\\\\ &\ \ (12)\ \lim_{x \rightarrow 0}x^2e^{\frac{1}{x^2}}=\lim_{x \rightarrow 0}\frac{e^{\frac{1}{x^2}}}{\frac{1}{x^2}}=\lim_{x \rightarrow 0}e^{\frac{1}{x^2}}=+\infty\\\\ &\ \ (13)\ \lim_{x \rightarrow 1}\left(\frac{2}{x^2-1}-\frac{1}{x-1}\right)=\lim_{x \rightarrow 1}\frac{2-(x+1)}{(x+1)(x-1)}=\lim_{x \rightarrow 1}\frac{-1}{x+1}=-\frac{1}{2}\\\\ &\ \ (14)\ 令\frac{1}{t}=\frac{a}{x}，则x=ta，当x \rightarrow \infty时，t \rightarrow \infty，\\\\ &\ \ \ \ \ \ \ \ \ \ \lim_{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^x=\lim_{t \rightarrow \infty}\left(1+\frac{1}{t}\right)^{ta}=\left[\lim_{t \rightarrow \infty}\left(1+\frac{1}{t}\right)^t\right]^a=e^a\\\\ &\ \ (15)\ \lim_{x \rightarrow 0^+}x^{sin\ x}=e^{{\displaystyle \lim_{x \rightarrow 0^+}sin\ xln\ x}}，而\lim_{x \rightarrow 0^+}sin\ xln\ x=\lim_{x \rightarrow 0^+}\frac{sin\ x}{x} \cdot \frac{ln\ x}{\frac{1}{x}}=\lim_{x \rightarrow 0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x \rightarrow 0^+}(-x)=0，\\\\ &\ \ \ \ \ \ \ \ \ \ \ 所以\lim_{x \rightarrow 0^+}x^{sin\ x}=e^0=1\\\\ &\ \ (16)\ \lim_{x \rightarrow 0^+}\left(\frac{1}{x}\right)^{tan\ x}=e^{{\displaystyle \lim_{x \rightarrow 0^+}tan\ xln\ \frac{1}{x}}}，而\lim_{x \rightarrow 0^+}tan\ xln\ \frac{1}{x}=\lim_{x \rightarrow 0^+}\frac{tan\ x}{x}\cdot \frac{-ln\ x}{\frac{1}{x}}=\lim_{x \rightarrow 0^+}\frac{-\frac{1}{x}}{-\frac{1}{x^2}}=0，\\\\ &\ \ \ \ \ \ \ \ \ \ \ 所以\lim_{x \rightarrow 0^+}\left(\frac{1}{x}\right)^{tan\ x}=e^0=1 & \end{aligned} ​  (1) x→0lim​xln(1+x)​=x→0lim​1+x1​=1  (2) x→0lim​sin xex−e−x​=x→0lim​cos xex+e−x​=2  (3) x→0lim​x−sin xtan x−x​=x→0lim​1−cos xsec2 x−1​=x→0lim​cos2 x1+cos x​=2  (4) x→πlim​tan 5xsin 3x​=x→πlim​5sec2 5x3cos 3x​=x→πlim​53​cos 3x⋅cos2 5x=−53​  (5) x→2π​lim​(π−2x)2ln sin x​=x→2π​lim​8x−4πsin xcos x​​=x→2π​lim​8x−4πcot x​=x→2π​lim​8−csc2 x​=−81​  (6) x→alim​xn−anxm−am​=x→alim​nxn−1mxm−1​=x→alim​nm​xm−n=nm​am−n  (7) x→0+lim​ln tan 2xln tan 7x​=x→0+lim​tan 2x2sec2 2x​tan 7x7sec2 7x​​=x→0+lim​2sin 14x7sin 4x​=x→0+lim​28cos 14x28cos 4x​=1  (8) x→2π​lim​tan 3xtan x​=x→2π​lim​3sec2 3xsec2 x​=x→2π​lim​3cos2 xcos2 3x​=x→2π​lim​−6cos xsin x−6cos 3xsin 3x​=x→2π​lim​sin 2xsin 6x​=x→2π​lim​2cos 2x6cos 6x​=3  (9) x→+∞lim​arccot xln(1+x1​)​=x→+∞lim​−1+x21​−x+x21​​=x→+∞lim​1+x1​1+x21​​=1  (10) x→0lim​sec x−cos xln(1+x2)​=x→0lim​sin xtan xln(1+x2)​=x→0lim​cos xtan x+sin xsec2 x1+x22x​​=x→0lim​sin x+tan xsec x1+x22x​​=          x→0lim​cos x+sec3 x+sec xtan2 x(1+x2)22(1−x2)​​=1  (11) x→0lim​xcot 2x=x→0lim​sin 2xxcos 2x​=x→0lim​2cos 2xcos 2x−2xsin 2x​=x→0lim​(21​−xtan 2x)=21​  (12) x→0lim​x2ex21​=x→0lim​x21​ex21​​=x→0lim​ex21​=+∞  (13) x→1lim​(x2−12​−x−11​)=x→1lim​(x+1)(x−1)2−(x+1)​=x→1lim​x+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+lim​xsin x=ex→0+lim​sin xln x，而x→0+lim​sin xln x=x→0+lim​xsin x​⋅x1​ln x​=x→0+lim​−x21​x1​​=x→0+lim​(−x)=0，           所以x→0+lim​xsin x=e0=1  (16) x→0+lim​(x1​)tan x=ex→0+lim​tan xln x1​，而x→0+lim​tan xln x1​=x→0+lim​xtan 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 \begin{aligned} &\ \ \lim_{x \rightarrow \infty}\frac{x+sin\ x}{x}=\lim_{x \rightarrow \infty}\left(1+\frac{sin\ x}{x}\right)=1 & \end{aligned} ​  x→∞lim​xx+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 \begin{aligned} &\ \ \lim_{x \rightarrow 0}\frac{x^2sin\ \frac{1}{x}}{sin\ x}=\lim_{x \rightarrow 0}\frac{xsin\ \frac{1}{x}}{\frac{sin\ x}{x}}=0 & \end{aligned} ​  x→0lim​sin xx2sin x1​​=x→0lim​xsin x​xsin 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。 \begin{aligned} &\ \ \lim_{x \rightarrow 0^+}f(x)=\lim_{x \rightarrow 0^+}\left[\frac{(1+x)^{\frac{1}{x}}}{e}\right]^{\frac{1}{x}}=e^{{\displaystyle \lim_{x \rightarrow 0^+}\frac{1}{x}ln\left[\frac{(1+x)^{\frac{1}{x}}}{e}\right]}}，\\\\ &\ \ 而\lim_{x \rightarrow 0^+}\frac{1}{x}ln\left[\frac{1}{x}ln(1+x)-1\right]=\lim_{x \rightarrow 0^+}\frac{ln(1+x)-x}{x^2}=\lim_{x \rightarrow 0^+}\frac{\frac{1}{1+x}-1}{2x}=\lim_{x \rightarrow 0^+}-\frac{1}{2(1+x)}=-\frac{1}{2}，所以\lim_{x \rightarrow 0^+}f(x)=e^{-\frac{1}{2}}，\\\\ &\ \ 因\lim_{x \rightarrow 0^-}f(x)=\lim_{x \rightarrow 0^-}e^{-\frac{1}{2}}=e^{-\frac{1}{2}}，f(0)=e^{-\frac{1}{2}}，所以\lim_{x \rightarrow 0^+}f(x)=\lim_{x \rightarrow 0^-}f(x)=f(0)，函数f(x)在x=0处连续。 & \end{aligned} ​  x→0+lim​f(x)=x→0+lim​[e(1+x)x1​​]x1​=ex→0+lim​x1​ln[e(1+x)x1​​]，  而x→0+lim​x1​ln[x1​ln(1+x)−1]=x→0+lim​x2ln(1+x)−x​=x→0+lim​2x1+x1​−1​=x→0+lim​−2(1+x)1​=−21​，所以x→0+lim​f(x)=e−21​，  因x→0−lim​f(x)=x→0−lim​e−21​=e−21​，f(0)=e−21​，所以x→0+lim​f(x)=x→0−lim​f(x)=f(0)，函数f(x)在x=0处连续。​​