Calculate #lim_(xto1)(e^f(x)-1)/(x-1)# ?

Question

Given #f:RR->RR# continuous at #x_0=1# with #lim_(xto1)f(x)/(x-1)=4#

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Answer 1 · 2018-06-10T07:52:30

Deleted (wrong answer)

Answer 2 · 2018-06-10T08:39:54

Any good?

#lim_(xto1)(e^f(x)-1)/(x-1)#

#= lim_(xto1)((1 + f(x) + (f^2(x))/(2!) + ...)-1)/(x-1)#

#= lim_(xto1)( f(x) + (f^2(x))/(2!) + ...)/(x-1)#

#= lim_(xto1)( f(x))/(x-1) + lim_(xto1) ((f^2(x))/(2!) )/(x-1) + ...#

#= lim_(xto1)( f(x))/(x-1) + lim_(xto1) (f(x))/(2!) f(x)/(x-1) + ...#

#= 4+ 0* 4 + ... = 4#

Answer 3 · 2018-06-10T09:54:23

For #x# near #1# , consider

#{(g(x)=f(x)/(x-1)" "),(lim_(xto1)g(x)=4" "):}# #<=>#

#{(f(x)=(x-1)g(x)" "),(lim_(xto1)g(x)=4" "):}#

Hence, #f(1)=lim_(xto1)f(x)=lim_(xto1)(x-1)g(x)=4*0=0#

#I=lim_(xto1)(e^f(x)-1)/(x-1)=lim_(xto1)(e^((x-1)g(x))-1)/(x-1)#

#lim_(xto1)g(x)=4# so there is #a>0# for which we have #0<##a<##g(x)# when #x->1#

#=# #lim_(xto1)(e^((x-1)g(x))-1)/((x-1)*g(x))*g(x)#

Substitute

#(x-1)g(x)=t#
#x->1#
#t->0#

#lim_(t->0)(e^t-1)/t=_(DLH)^((0/0))lim_(t->0)((e^t-1)')/((t)')#

#=lim_(t->0)e^t=e^0=1#

Therefore #I=lim_(xto1)(e^f(x)-1)/(x-1)=1*4=4#