Prove that $\lim\limits_{x \to 1} \frac{x+2}{x^2+1} = \frac{3}{2}$ I am required to prove the following limit using the epsilon-delta definition:
$$\lim_{x\to 1}\frac{x+2}{x^2+1}=\frac{3}{2}$$
So, ($\forall\epsilon>0)(\exists\delta>0)[0<\vert x-1\vert<\delta \implies\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert<\epsilon]$
Below is my working on getting the $\delta$ that I need:
$\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert\le\vert\frac{x+2}{x^2+1}\vert+\frac{3}{2}=\frac{\vert x+2 \vert}{\vert x^2+1 \vert}+\frac{3}{2}$ (Triangle Inequality)
But I know that $x^2+1\ge1$ for all values of $x$, so I can say that $\frac{\vert x+2 \vert}{\vert x^2+1 \vert}+\frac{3}{2}\le\vert x-1+3\vert+\frac{3}{2}\le \vert x-1\vert +3 + \frac{3}{2}<\delta+\frac{9}{2}$
So for every $\epsilon$, I choose $\delta=\epsilon-\frac{9}{2}$, then:$$0<\vert x-1\vert<\delta \implies\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert \le \vert\frac{x+2}{x^2+1}\vert+\frac{3}{2}\le\vert x+2\vert+\frac{3}{2}\le\vert x-1+3\vert+\frac{3}{2}\le \vert x-1\vert +3 + \frac{3}{2}<\delta+\frac{9}{2}=\epsilon$$
Am I doing this right? Is there something wrong to assume that $x^2+1\ge1$ for all values of $x$?
 A: Hint:)
Notice that you can't never conclude from 
$$|f(x)-\ell|\leq|f(x)|+|\ell|<\epsilon~~~~,~~~~~\ell\neq0$$
obtained by triangle inequality, a limit yield.
A: $\large{\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert=\vert\frac{2x+4-3x^2-3}{2(x^2+1)}\vert=\vert\frac{(x-1)(-3x-1)}{2(x^2+1)}\vert}$ Can you take it further?
A: $|\frac{x+2}{x^2 +1} - \frac{3}{2}| = |\frac{-3x^2 + 2x + 1}{2(x^2+1)}|$
Using your same logic about $x^2 + 1$, we can bound it from above this way:
$|\frac{-3x^2 + 2x + 1}{2(x^2+1)}| \leq |\frac{-3x^2+2x+1}{2}| \leq |-3x^2+2x+1| = |3x^2-2x-1| = |(3x+1)(x-1)|$
We're looking for any $\delta > 0$ that will make this expression less than $\epsilon$. Our main problem here is that we need a way to bound the expression $3x+1$. We can do this by letting $\delta \leq 1$. This is allowed because even if we can find a $\delta > 1$ that works, any $\delta$ less than it will work as well. So, if $\delta \leq 1$ then:
$|x-1| < 1 \implies -1 < x-1 < 1 \implies 0 < x < 2 \implies 1 < 3x+1 < 7$
Now we can bound our limit expression from above. If $\delta \leq 1$ and $|x-1| < \delta$ then:
$|(3x+1)(x-1)| < 7 \delta$
Which is less than epsilon if $\delta =\frac{\epsilon}{7}$
You might think you're done, and $\delta = \frac{\epsilon}{7}$. But remember our solution depended on $\delta \leq 1$. So what if $\epsilon = 100$? For this reason, we need to take the minimum of 1 and $\frac{\epsilon}{7}$. So the final answer is $\delta = \min\{\frac{\epsilon}{7}, 1\}$
