Limits and Derivatives · Question 2

Let x0 be the real number such that ex0+x0=0. For a given real number , define g(x)= 3xex+3x- ex- x3(ex+1) for all real numbers x. Then which one of the following statements is TRUE?

1. For =2, x x0 | g(x)+ex0x-x0 |=0
2. For =2, x x0 | g(x)+ex0x-x0 |=1
3. For =3, x x0 | g(x)+ex0x-x0 |=0
4. For =3, x x0 | g(x)+ex0x-x0 |= 23

Answer

(3) For =3, x x0 | g(x)+ex0x-x0 |=0

Solution

Write g(x)=x- 3 ex+xex+1.

At x0: ex0+x0=0, so g(x0)=x0=-ex0. Hence g(x)+ex0=g(x)-g(x0) and the limit is |g'(x0)|.

g'(x)=1- 3 (ex+1)2-(ex+x)ex(ex+1)2; at x0 the term (ex0+x0)=0, so g'(x0)=1- 3.

=2: limit = 13 ; =3: limit =0.

Answer: (C)