Vtyjkyuk

I need to either compute $$lim_xto 0 dfrac1 + 1/x1 + 1/x^2$$ using definition of a limit or prove it doesn't exist.

My attempt:

Given $epsilon > 0$, we wish to find $N$ such that $forall x geq N$,

$$left|dfrac1 + 1/x1 + 1/x^2 - 0right| < epsilon.$$

We have

$$left|dfrac1 + 1/x1 + 1/x^2 - 0right| = left|dfracx^2 + xx^2 + 1right| leq dfracx^2 + xx = x + 1 < epsilon. $$

Thus, we can choose $N geq epsilon - 1$ and then whenever $x > N$, we have $left|f(x)right| < epsilon implies lim_xto 0 f(x) = 0$

Let $f:mathbbRsetminus0to mathbbR$ be defined by
$$f(x)=frac1 + 1/x1 + 1/x^2$$

Claim:$;lim_xto 0f(x) = 0$.

Let $epsilon > 0$, and let $delta=minbigl(1,largefracepsilon2bigr)$.

Suppose $|x| < delta$, $xne 0$. We want to show that $|f(x)| < epsilon$.

Since $|x| < 1$, it follows that $|x+1| < 2$, hence
beginalign*
|f(x)|
&=left|frac1 + 1/x1 + 1/x^2right|\[4pt]
&=left|fracx^2+xx^2 + 1right|\[4pt]
&=fracleftleft\[4pt]
& < left|x^2+xright|\[4pt]
&=left|x(x+1)right|\[4pt]
&=|x||x+1|\[4pt]
&< 2left|xright|\[4pt]
& < 2delta\[4pt]
&< epsilon\[4pt]
endalign*
as was to be shown.

You seem to be mixing two different $epsilon$-$delta$ definitions. Here they are:

1. Let $ x_n _n in mathbbN$ be a sequence in $mathbbR$. We say that $x_n$ converges to $x$ if for every $epsilon > 0$ there exists $N in mathbbN$ such that $|x_n - x| < epsilon$ for all $n geq N$.


2. Let $f : mathbbR to mathbbR$ and $c in mathbbR$. We say that $f$ has limit $L$ at $c$ if for every $epsilon > 0$ there exists $delta > 0$ such that $|f(x) - L| < epsilon$ whenever $0 < |x-c| < delta$.

You need to be using the second definition, but somehow there is an $N$ in your proof, and you are making a statement for all $x geq N$, which doesn't make sense.

However, the guess of $L = 0$ is accurate, and we have
$$
left| frac1+1/x1+1/x^2 - 0 right| = left| fracx^2 + xx^2 + 1 right|.
$$
Therefore, if $0 < |x| < delta$ for small enough delta (say, $0 < delta < 1$), then
$$
left| fracx^2 + xx^2 + 1 right| = |x|fracx+1x^2 + 1 < delta(delta + 1),
$$
where the last inequality holds because $inf < delta = 0$, and it is not attained inside the domain. Thus, given $epsilon > 0$ if we choose $delta > 0$ such that $delta < 1$ and $delta(delta + 1) < epsilon$ then we are done. The unique positive root of $y^2 + y - epsilon = 0$ is
$$
frac-1 + sqrt1 + 4epsilon2.
$$
Let $epsilon_1 = min epsilon, 1 $, and choose
$$
delta = frac-1 + sqrt1 + 4epsilon_12.
$$
Then, this choice of $delta$ satisfies our required criteria.