Finding the Limit of a Function as x Approaches Negative Infinity

[John O' Meara]

Find a negative number N such that $$|\frac{x^2}{1+x^2}-1| < \epsilon$$ for x < N. That implies $$|\frac{-1}{1+x^2}| < \epsilon if x < N$$.This gives me the following $$-|\frac{1}{1+x^2}| < \epsilon if x < N$$.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 $$|\frac{x^2}{1+x^2}-1| < \epsilon$$ for x < N. That implies $$|\frac{-1}{1+x^2}| < \epsilon if x < N$$.This gives me the following $$-|\frac{1}{1+x^2}| < \epsilon if x < N$$.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 $|\frac{-1}{1+x^2}|< epsilon$" is more "standard" and gives N= -M. Saying that $|\frac{-1}{1+x^2}|< \epsilon$ is the same as saying $0< \frac{1}{1+x^2}< \epsilon$ which is, in turn, the same as saying that $1+ x^2> \frac{1}{\epsilon}$ which is the same as $x^2> \frac{1}{\epsilon}- 1$. 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
$$\lim_{x-> -\infty}f(x)=L$$
if given any number $$\epsilon >0$$, there corresponds a negative number N such that
$$|f(x)-L| < \epsilon \mbox{ if } x < N$$.