Prove that $\lim\limits_{x \to 1} \frac{x+2}{x^2+1} = \frac{3}{2}$

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.

$\large{\vert\frac{x+2}{x^2+1}-\frac{3}{2}\vert=\vert\frac{2x+4-3x^2-3}{2(x^2+1)}\vert=\vert\frac{(x-1)(-3x-1)}{2(x^2+1)}\vert}$ Can you take it further?

$|\frac{x+2}{x^2 +1} - \frac{3}{2}| = |\frac{-3x^2 + 2x + 1}{2(x^2+1)}|$

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 $\delta = \min\{\frac{\epsilon}{7}, 1\}$