Use the $\epsilon$-$\delta$ definition to prove the following limit

Question

Here is the question I am trying to solve: $\lim_{x\rightarrow0}\frac{x^2+1}{x+1}=1$

Here is what I have so far:

Proof

Let $\epsilon>0$. We will show given $\delta\geq0$, we have $0<|x|<\delta$ implies that

$$\lvert\frac{x^2+1}{x+1}-1\lvert = \lvert\frac{x^2+1-x-1}{x+1}\lvert = \lvert\frac{x^2-x}{x+1}\lvert = \frac{|x||x-1|}{|x+1|}$$

But this is where I get stuck. I am unsure what to do to make sure the denominator is bounded away from $0$. Any help welcome!

Answer 1

Following on the hints, notice that $|x+1|\geq 0$, and we have equality at $-1$, So if $x=0$ we have the denominator is $1$, if $x>0$ the denominator is $|x+1|>1$. The key is to bound away from $-1$ so we can get rid of the absolute value sign.

So as long as $x>-1$ we have $|x+1|=x+1>0$, in particular if $\delta<1$ then all of our points will be in this area

So, we know we need $\delta$ to be the minimum of $1$ and some bound based on $\epsilon$. Given that,we know the lowest $x$ can be is $1-\delta$, which means that's an upper bound for the value in the denominator:

$$\frac {|x||x-1|}{|x+1|}=\frac {|x||x-1|}{x+1}\leq \frac {|x||x-1|}{1-\delta }$$ $|x|<\delta$, and in our range near 0, $|x-1|=1-x$, which is minimized at $x=-\delta$, so we can bound the numerator by how small it can get:

$$\frac {|x||x-1|}{1-\delta }\leq \frac {\delta (1+\delta) }{1-\delta }$$ Now you just need to do a bit of polynomial division to figure out how small $\delta$ has to be in terms of $\epsilon$ to get this expression less than $\epsilon$

Answer 2

Take any $\epsilon>0$, take $0<\delta<\min\{\frac{1}{2},\frac{\epsilon}{3}\}$. When $0<|x-0|<\delta$, we have (by your computation) $$\left|\frac{x^2+1}{x+1}-1\right|=|x|\left|\frac{x-1}{x+1}\right|.$$ Now, note that $\left|\frac{x-1}{x+1}\right|<3$, we have $$|x|\left|\frac{x-1}{x+1}\right|<3\delta<\epsilon$$ as desired.

Remark:

1. $\left|\frac{x-1}{x+1}\right|<3$ because $0<|x|<\delta<\frac{1}{2}$.
2. $3\delta<\epsilon$ because $\delta<\frac{\epsilon}{3}$.

Answer 3

Let $\epsilon >0$; we are searching for a $δ>0$, such that $x\in (-δ,δ)\setminus \{0\}\Rightarrow |\frac {x^2+1}{x+1}-1|<\epsilon\iff |\frac {x^2-x}{x+1}|<\epsilon\iff \frac {|x||x-1|}{|x+1|}<\epsilon$ (1). But, for $δ\in (0,\frac 12]$, $x\in (-δ,δ)\setminus \{0\}$ gives $x\in (-\frac 12,\frac 12)\setminus \{0\}$, so $x-1\in (-\frac 32,-\frac 12)\setminus \{-1\}\Rightarrow \frac 12<|x-1|<\frac 32, x\neq 0$ and $x+1\in (\frac 12,\frac 32)\setminus \{1\}\Rightarrow \frac 12<|x+1|<\frac 32, x\neq 0$. Inspired from (1) we demand $\frac {|x||x-1|}{|x+1|}<\frac {δ\frac 32}{\frac 23}\leq\epsilon\Rightarrow 0<δ\leq \frac 49 \epsilon$. By choosing $0< δ\leq min\{\frac 12, \frac 49\epsilon\}$ we have $x\in (-δ,δ)\setminus \{0\}\Rightarrow |\frac {x^2+1}{x+1}-1|=\frac {|x||x-1|}{|x+1|}<\frac {δ\frac 32}{\frac 23}\leq \frac {\frac 49 \epsilon \frac 32}{\frac 23}=\epsilon$. q.e.d.