As the title suggest I have to prove a limit:
$$\lim_{x\to2} \frac{x-2}{1+x^2}=0$$
it is the denominator which throws me a of a little since I get a variable whilst trying to find a relationship between epsilon and small delta.
I get $$\left|\frac{x-2}{1+x^2} - 0\right| < \epsilon$$
which gives me $|x-2| < \epsilon(1+x^2)$
i.e $\delta = \epsilon(1+x^2)$
Where do I go from this?? Thank you in advance!
On
For questions like this (when you can't factor) you have to bound things. So for
For $ x \in (1,3)$, $($i.e. $|x-2|<1)$, $\space 2 < x^2+1<10$.
In particular,
$$|x-2|<1 \implies 2 < |x^2+1| \implies \frac{1}{|x^2+1|} < \frac{1}{2}$$
Set $\delta = min\{1,\epsilon\}$
Then $$|x-2|<\delta \implies |x-2|<1,\epsilon \implies \bigg|\frac{x-2}{x^2+1}\bigg| = |x-2| * \frac{1}{|x^2+1|} < \epsilon * 1/2 = \epsilon /2 < \epsilon$$
EDIT-In response to your comment below:
I'm not setting $\delta = 1$ I'm setting it whichever of $ \epsilon,1$ is smaller. This depends on what $\epsilon$ is. Lets consider both cases. If $\epsilon >1$ then $\delta = min\{\epsilon, 1\} = 1$
As mentioned above if $|x-2| < 1$ then $$\frac{1}{|x^2+1|} < \frac{1}{2}$$
Thus,
$$|x-2| < \delta \implies \bigg|\frac{x-2}{x^2+1}\bigg| = |x-2| * \frac{1}{|x^2+1|} < \delta* \frac{1}{2} = \frac{\delta}{2} = \frac{1}{2} < 1 \space < \epsilon$$
If on the other hand $\epsilon < 1$, then
$$\delta = min\{1,\epsilon\} = \epsilon $$
Since $\delta = \epsilon < 1$ we still have
$$|x-2| < \delta \implies |x-2| < 1 \implies \frac{1}{|x^2+1|} < \frac{1}{2} $$
and hence
$$|x-2| < \delta \implies \bigg|\frac{x-2}{x^2+1}\bigg| = |x-2| * \frac{1}{|x^2+1|} < \delta* \frac{1}{2} = \frac{\delta}{2} = \frac{\epsilon}{2} < \space \epsilon$$
The idea behind this is that when working with limits all that matters is the behavior of the function NEAR the limit point. So although sometimes we can't bound the absolute value of something altogether, we can bound it "near" the point in question. In this instance, we were able to bound $$\frac{1}{|x^2+1|}$$ by $1/2$ when $x$ was near 2
As David Reed pointed out, the choice of $\delta$ cannot depend on $x$. What you should do is form an estimate for the $\frac{1}{x^2 + 1}$ term. If you think about it, this term has a maximum value: $1$. So, we always have, $$\left|\frac{x - 2}{x^2 + 1}\right| = \frac{|x - 2|}{x^2 + 1} \le |x - 2| \cdot 1.$$ Thus, you can choose $\delta = \varepsilon$.
EDIT: To see why $\frac{1}{x^2 + 1} \le 1$, first note that $x^2 \ge 0$, hence $x^2 + 1 \ge 1$, so $\frac{1}{x^2 + 1} \le 1$.
Why is this useful? As I said, it allows us to set $\delta = \varepsilon$, for, $$0 < |x - 2| < \varepsilon \implies \left|\frac{x - 2}{x^2 + 1} - 0\right| \le |x - 2| < \varepsilon,$$ completing the proof.
More generally, it's really nice in an $\varepsilon$-$\delta$ proof that $\lim_{x\to a} f(x) = L$ if you can reduce $|f(x) - L|$ down to $|x - a| \cdot g(x)$, where $g(x)$ is globally bounded above. In our case, $g(x) = \frac{1}{x^2 + 1}$. Then, if $g(x) \le M$ for some $M > 0$, then $$0 < |x - a| < \frac{\varepsilon}{M} \implies |f(x) - L| = |x - a|\cdot g(x) < \frac{\varepsilon}{M} \cdot M = \varepsilon.$$ It means that this unpredictable factor of $g(x)$ is only $M$ at its worst. We are trying to make $|f(x) - L|$ very small, controlling only $|x - a|$, and we can't have large values of $g(x)$ spoiling this. But since $g(x)$ only gets $M$ large, we can make $|x - a|$ smaller to compensate for this.
When $g(x)$ is not globally bounded above (e.g. if it has an asymptote), that's when you have to bound things, like David Reed did. Let's take an example where this is necessary. Let's prove that $\lim_{x \to 2} \frac{x - 2}{x - 3} = 0$.
In this case, $g(x) = \frac{1}{|x - 3|}$, which is certainly not bounded above! It has an asymptote at $x = 3$, which is bad for us, since $g(x)$ will be getting very large around $x = 3$, while we're making $|x - 2|$ small. The good news is, when $|x - 2|$ is small enough, we can force $x$ away from this bad behaviour. If we force $|x - 2| < \frac{1}{2}$, then $x$ is bounded away from the bad behaviour at $x = 3$.
While we were not be able to bound $g(x)$ globally, what we can do is bound it locally when $|x - 2| < \frac{1}{2}$, i.e. when $x \in \left(\frac{3}{2}, \frac{5}{2}\right)$. Note that for $x$ in this range, we have $x < 3$ so $|x - 3| = 3 - x$. Note that $g(x) = \frac{1}{3 - x}$ is an increasing function on this interval, hence for $x$ in the interval, $$\frac{2}{3} = g\left(\frac{3}{2}\right) \le g(x) \le g\left(\frac{5}{2}\right) = 2.$$ In particular, when $x \in \left(\frac{3}{2}, \frac{5}{2}\right)$, we have $g(x) \le 2$. Again it's not a global upper bound, but it's a local upper bound, and that's good enough for us. Then, \begin{align*} 0 < |x - 2| < \min\left\lbrace \frac{1}{2}, \frac{\varepsilon}{2}\right\rbrace &\implies |x - 2| < \frac{1}{2} \text{ and } 2|x - 2| < \varepsilon \\ &\implies \frac{1}{|x - 3|}|x - 2| < \varepsilon \\ &\implies \left|\frac{x - 2}{x - 3} - 0 \right| < \varepsilon. \end{align*} It's worth noting that while my choice of $\frac{1}{2}$ was arbitrary, I couldn't choose $1$ or anything higher, as this would allow $g(x)$ to become unbounded. The whole point is to force $x$ away from these problem spots, to allow us to bound $g(x)$.