Thread: Prove left-hand limit against neg. infinity

1. Prove left-hand limit against neg. infinity

Hello, I need some help with this:

Prove: lim x --> -1- 5/(x+1)^3 = neg. infinity

I am not sure how to do that.
a=-1

For a limit against neg. infinity I know that
if 0 < |x-a| < delta then f(x) < N

and for a left-hand limit I know
if a-delta < x < a then |f(x) - L| < epsilon

So I have
5/(x+1)^3 < N if |x+1| < delta
and
-1-delta < x < -1 if ???

I do not know how to combine them.
Thanks for any hints..

2. Originally Posted by DBA
Hello, I need some help with this:

Prove: lim x --> -1- 5/(x+1)^3 = neg. infinity

I am not sure how to do that.
a=-1

For a limit against neg. infinity I know that
if 0 < |x-a| < delta then f(x) < N

and for a left-hand limit I know
if a-delta < x < a then |f(x) - L| < epsilon

So I have
5/(x+1)^3 < N if |x+1| < delta
and
-1-delta < x < -1 if ???

I do not know how to combine them.
Thanks for any hints..
Did you try graphing the function.

If you graph it, it's really quite obvious...

3. Hello, I did graph it, but for me nothing is obvious. I need to proof it without a graph and I have no clue how...

4. Originally Posted by DBA
Hello, I did graph it, but for me nothing is obvious. I need to proof it without a graph and I have no clue how...
What happens to $y$ as you approach $x = -1$ from the left?
If I remember correctly, $\epsilon \delta$ proofs are only used when you are trying to find a whole limit, not a left or right-hand limit... It's been a while since I've done $\epsilon \delta$ proofs though...
5. If x< -1 then x+1< 0 so $(x+ 1)^3< 0$ and $\frac{5}{(x+1)^3}< 0$. As x approaches -1, the denominator approaches 0 while the numerator stays 5- ignoring the sign for the moment, that fraction gets larger and larger. Because it is negative, the fraction goes to $-\infty$.