# Thread: piecewise limits

1. ## piecewise limits

1.
Let y={x+1 if x>=1
{-x^2 if x<1
Does lim x->1 f(x) exists?

2.
y={ax+1 if x>=1
{-x^2 if x<1
Find all values of a so that lin x->1 f(x) exists

2. Originally Posted by bobby77
1.
Let y={x+1 if x>=1
{-x^2 if x<1
Does lim x->1 f(x) exists?
For this limit to exist requires that

lim(x->1+) f(x) = lim(x->1-) f(x)

and both be finite (lim(x->1+) means the limit as x goes to 1 from above,
and lim(x->1-) means the limit as x goes to 1 from below.]

Now:

lim(x->1+) f(x) = lim(x->1+) x+1 = 2

and

lim(x->1-) f(x) = lim(x->1-) -x^2 = -1.

Hence:

lim(x->1+) f(x) != lim(x->1-) f(x)

and so the limit does not exist.

RonL

3. Originally Posted by bobby77
1.
2.
y={ax+1 if x>=1
{-x^2 if x<1
Find all values of a so that lin x->1 f(x) exists
As before for the limit to exist requires that

lim(x->1+) f(x) = lim(x->1-) f(x)

and both be finite (lim(x->1+) means the limit as x goes to 1 from above,
and lim(x->1-) means the limit as x goes to 1 from below.]

Now:

lim(x->1+) f(x) = lim(x->1+) ax+1 = a+1

and

lim(x->1-) f(x) = lim(x->1-) -x^2 = -1.

Hence:

lim(x->1+) f(x) = lim(x->1-) f(x)

only if a+1=-1, which is equivalent to a=-2.

RonL