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

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

3 Answers
Jun 10, 2018

Deleted (wrong answer)

Jun 10, 2018

Any good?

Explanation:

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) + ...

  • lim_(xto1)f(x)/(x-1)=4 implies lim_(xto1)f(x) =0

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

Jun 10, 2018

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)

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