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Exercise 3-2, Volume 1, Tongji Fifth Edition, Advanced Mathematics The fourth question I calculated is E- 1, so it is discontinuous.
lnf(x)=( 1/x)[( 1/x)ln( 1+x)- 1]=[ln( 1+x)-x]/x^2

lim(x→0+)[ln( 1+x)-x]/x^2=lim(x→0+){[ 1/( 1+x)]- 1}/(2x)=lim(x→0+)[-x/(2x)]lim(x→0+)[ 1/( 1+x)]=- 1/2

∴lim(x→0+)f(x)=e^(- 1/2)

And lim (x → 0-) f (x) = f (0) = e (-1/2).

∴f(x) is continuous at x=0.

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