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

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- Oct 6th 2006, 12:22 PMbobby77piecewise 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 - Oct 6th 2006, 12:33 PMCaptainBlack
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 - Oct 6th 2006, 12:36 PMCaptainBlack
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