<s>[INST] 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 $x$?
 [/INST] 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.
</s>