Finding the Limit of a Function as x Approaches Negative Infinity

[John O' Meara]

Find a negative number N such that |\frac{x^2}{1+x^2}-1| &lt; \epsilon for x < N. That implies |\frac{-1}{1+x^2}| &lt; \epsilon if x &lt; N.This gives me the following -|\frac{1}{1+x^2}| &lt; \epsilon if x &lt; N.I do not know what to do from here. Please help as I am teaching myself. Thanks.

 

[HallsofIvy]

Find a negative number N such that |\frac{x^2}{1+x^2}-1| &lt; \epsilon for x < N. That implies |\frac{-1}{1+x^2}| &lt; \epsilon if x &lt; N.This gives me the following -|\frac{1}{1+x^2}| &lt; \epsilon if x &lt; N.I do not know what to do from here. Please help as I am teaching myself. Thanks.

Finding "a negative number N such that if x< N then ..." is very peculiar. But since x only occurs to even power, it doesn't matter whether you use x or -x. "Find a positive number M such that if x> N then |\frac{-1}{1+x^2}|&lt; epsilon" is more "standard" and gives N= -M. Saying that |\frac{-1}{1+x^2}|&lt; \epsilon is the same as saying 0&lt; \frac{1}{1+x^2}&lt; \epsilon which is, in turn, the same as saying that 1+ x^2&gt; \frac{1}{\epsilon} which is the same as x^2&gt; \frac{1}{\epsilon}- 1. Can you carry on from there?

 

[John O' Meara]

x=+/-\sqrt{\frac{1-\epsilon}{\epsilon}} Then I have to determine which one of the roots is the answer, as x is negitive it must -\sqrt{\frac{1-\epsilon}{\epsilon}}, but that last answer doesn't sound a very good reason as to why it is the negitive root. I was wondering is there better reasoning?

 

[John O' Meara]

The definition I am using is: Let f(x) be defined for all x in some infinite open interval extending in the negative x-direction. We will write
\lim_{x-&gt; -\infty}f(x)=L
if given any number \epsilon &gt;0, there corresponds a negative number N such that
|f(x)-L| &lt; \epsilon \mbox{ if } x &lt; N.