I've been trying to proven: [math]\lim\limits_{x \rightarrow 1} \frac{x}{x^{2} + 1} = \frac{1}{2}[/math]
Using just the following definition: [math]\lim\limits_{x \rightarrow a} f(x) = L[/math] if for every [math]\varepsilon > 0[/math], there exits [math]\delta > 0[/math] such that [math]|f(x) - L| < \varepsilon[/math] whenever [math]0 < |x - a| < \delta[/math].
I'm really bad at factoring so I'm having a hard time finding a [math]\delta[/math]. I got to this point:
[math]\left|\cfrac{x}{x^{2} - 1} - \cfrac{1}{2}\right| = \left|\cfrac{x}{(x + 1)(x - 1)} - \cfrac{1}{2}\right| = \left|\cfrac{2x}{2(x + 1)(x - 1)} - \cfrac{(x + 1)(x - 1)}{2(x + 1)(x - 1)}\right| = \left|\cfrac{2x - (x + 1)(x - 1)}{2(x + 1)(x - 1)}\right|[/math].
Pretty sure it's shit though.. Can anyone help me with the algebra so I don't spend the rest of the night on one problem?
>>8005880
Is your denominator x^2 +1 or x^2 - 1?
>>8005900
[math]\frac{x}{x^{2}+1}[/math]
>>8005903
Do you see how your first step has an error?
>>8005913
oh yea.. but doesn't [math]x^{2} + 1 = (x + 1) \cdot (x - 1)[/math] anyway?
>>8005920
Multiply it out and tell me if it is the same thing.
>>8005922
I'm getting that you can pick delta to be the minimum of L and 1.
>>8005932
it should be the minimum of some epsilon and 1
>>8005937
Sorry, that is what I meant.
so now i'm here...
[math]\left| \cfrac{x}{x^{2} + 1} - \cfrac{1}{2} \right| = \left| \cfrac{2x}{2(x^{2} + 1)} - \cfrac{x^{2} + 1}{2(x^{2} + 1)} \right| = \left| \cfrac{2x - x^{2} - 1}{2(x^{2} + 1)} \right|[/math]
>>8005959
which
[math]\cfrac{|2x - x^{2} - 1|}{|2(x^{2} + 1)|} < \cfrac{|2x - x^{2} - 1|}{2}[/math]
>>8005978
you're so close
look at |x - a| < delta
what's your "a" in this situation?
>>8005959
[math]= \cfrac{|2x - x^{2} - 1|}{|2(x^{2} + 1)|} = \cfrac{|(x - 1)(1 - x)|}{|2(x^{2} + 1)|} < \cfrac{|x - 1|}{2} < \cfrac{\delta}{2} = \cfrac{2\eps}{2} = \eps \blacksquare[/math]
>>8005913
i probably would have spent a few more hours trying to figure out that problem with the first step being completely wrong...
>>8005880
if anyone is still around.. how do you prove something like this?
[math]\lim\limits_{x \rightarrow a} x^{n} = a^{n}[/math]
Seems like there is some trick to it....
>>8006076
induction
