Expert: Paul Klarreich - 4/30/2007

Question
I am stuck on a particular problem:

A function g is defined by the formula

g(x) = {x^2 + x + a,  x </= 0
      {((ln(1+x))/(sqrootx)), x > 0

For what values of "a" is g continuous at x = 0?

I don't even know how to attempt this.  My guess is that a would need to be 0, to get rid of "a", but that is a guess I don't know how to prove it.

Answer
Questioner:   Tiesh
Category:  Calculus
 
Subject:  Continuous Functions
Question:  I am stuck on a particular problem:

A function g is defined by the formula

g(x) = {x^2 + x + a,  x </= 0
     {((ln(1+x))/(sqrootx)), x > 0

For what values of "a" is g continuous at x = 0?

I don't even know how to attempt this.  My guess is that a would need to be 0, to get rid of "a",

but that is a guess I don't know how to prove it.
...............................................
Hi, Tiesh,

Definition of continuity:

f(x) is continuous at  x = a if:

1. f(a) exists.

2. lim(x->a) f(x) exists.

3. Those two are equal.

Now, for your 'multipart' function:

g(x) = {  x^2 + x + a,           x <= 0
      {  ((ln(1+x))/(sqrootx)), x > 0

The 'left part' is defined at  x = 0, because it says "x <= 0"  and

g(0) = a.

The 'right part' is not, but we can 'easily' compute

lim(x->0+) g(x)
      ln(1 + x)
lim    ----------
x->0   sqrt(x)

For this, I think we will use l'Hospital's rule, and differentiate top and bottom:
1/(1 + x)
------------- =
1/(2 sqrt(x))
2 sqrt(x)          0
--------- -->  ---------
1 + x              1

So the limit is zero.  
Therefore we require that  g(0) = 0, thus  a = 0.