jverkoey

10-24-2005, 07:11 PM

<edit>

Bah, after re-reading my post I guess it's not a very quick question. :rolleyes: </edit>

So I just took a midterm, and one of the questions was as follows:

lim (1-x^2)

x->+inf -------

(3+x)

I found two ways to go about doing this.

1) divide all by x:

lim (1-x^2)/x

x->+inf -------

(3+x)/x

lim (1/x-x)

x->+inf -------

(3/x+1)

When you plug in +inf, you get:

-inf

----

1

Evaluating to -inf.

I checked this over at the end of the test though, and decided I didn't like this solution, as by the definition of a limit: the left side limit must equal the right side limit...and how do you evaluate the left and right side of infinity?

So, I erased that answer and proceeded to write answer #2:

lim (1-x^2)

x->+inf -------

(3+x)

As lim lim

x->+inf 1/x->0.

Therefore,

lim (1-(1/x)^2)

1/x->0 ----------

(3+(1/x))

So, for lim 1/x->0+, we have:

-inf

----

+inf

Which is -inf.

Also, for lim 1/x->0-, we have:

-inf

----

-inf

Which is +inf.

lim1/x->0- DNE lim1/x->0+,

Therefore lim1/x->0 DNE.

So, yah, a bunch of my friends all said -inf is right, but I really don't like the idea of putting that down as saying it is -inf implies the limit exists and approaches -inf, which isn't necessarily true, as inf isn't defined as a number....Aurgh.

Bah, after re-reading my post I guess it's not a very quick question. :rolleyes: </edit>

So I just took a midterm, and one of the questions was as follows:

lim (1-x^2)

x->+inf -------

(3+x)

I found two ways to go about doing this.

1) divide all by x:

lim (1-x^2)/x

x->+inf -------

(3+x)/x

lim (1/x-x)

x->+inf -------

(3/x+1)

When you plug in +inf, you get:

-inf

----

1

Evaluating to -inf.

I checked this over at the end of the test though, and decided I didn't like this solution, as by the definition of a limit: the left side limit must equal the right side limit...and how do you evaluate the left and right side of infinity?

So, I erased that answer and proceeded to write answer #2:

lim (1-x^2)

x->+inf -------

(3+x)

As lim lim

x->+inf 1/x->0.

Therefore,

lim (1-(1/x)^2)

1/x->0 ----------

(3+(1/x))

So, for lim 1/x->0+, we have:

-inf

----

+inf

Which is -inf.

Also, for lim 1/x->0-, we have:

-inf

----

-inf

Which is +inf.

lim1/x->0- DNE lim1/x->0+,

Therefore lim1/x->0 DNE.

So, yah, a bunch of my friends all said -inf is right, but I really don't like the idea of putting that down as saying it is -inf implies the limit exists and approaches -inf, which isn't necessarily true, as inf isn't defined as a number....Aurgh.