Prove that $\lim_{x \rightarrow \infty} \frac{x^2+3}{4x^2-4x+8} = \frac{1}{4}$ using the $\epsilon$-$\delta $ definition of a limit.

Question

Prove that $$\lim_{x \rightarrow \infty} \frac{x^2+3}{4x^2-4x+8} = \frac{1}{4}$$ using the $\epsilon$-$\delta $ definition of a limit. So we want to find a $\delta$ such that for every $\epsilon>0$ $$x > \delta \rightarrow \left|\frac{x^2+3}{4x^2-4x+8} - \frac{1}{4}\right| = \left|\frac{x+1}{4x^2-4x+8}\right| < \epsilon. $$ The ratio has a global maximum at $x=2$ Thus, $x+1 < 2x$ and $4x^2-4x+8 > 2x^2 + 8 > 2x^2$. This implies that with $ \delta >2$ $$ x > \delta \rightarrow |\frac{x+1}{4x^2-4x+8}| < \frac{2x}{2x^2} = \frac{1}{x}. $$ I think this proves it... How do I wrap it up?

Answer 1

Note that for $x > 2$,

\begin{align}\tag{*}\left|\frac{x^2 + 3}{4x^2 - 4x + 8} - \frac{1}{4}\right| &= \left|\frac{x + 1}{4x^2 - 4x + 8}\right|\\ & = \frac{x + 1}{4(x^2 - x + 2)} \\ &< \frac{x + 1}{4(x - 1)^2}\\ & = \frac{1}{4(x - 1)} + \frac{1}{2(x - 1)^2} \\ &< \frac{3}{4(x - 1)}. \end{align}

Given $\epsilon > 0$, let $M = \max\{2,1 + \frac{3}{4\epsilon}\}$. If $x > M$, then the left-hand side of $(*)$ is less than $\frac{3}{4(x-1)}$, which is less than $\frac{3}{4(M-1)}$, which is less than $\epsilon$.

Answer 2

We observe: $$\begin{gathered} \frac{{{x^2} + 3}}{{4{x^2} - 4x + 8}} = \frac{1}{4} \cdot \frac{{{x^2} + 3}}{{{x^2} - x + 2}} \hfill \\ \hfill \\ \frac{1}{4} \cdot \frac{{{x^2} - x + 2 + x + 1}}{{{x^2} - x + 2}} = \frac{1}{4} \cdot (1 + \frac{{x + 1}}{{{x^2} - x + 2}}) = \frac{1}{4} + \frac{1}{4} \cdot \frac{{x + 1}}{{{x^2} - x + 2}} \hfill \\ \hfill \\ \frac{{{x^2} + 3}}{{4{x^2} - 4x + 8}} - \frac{1}{4} = \frac{1}{4} \cdot \frac{{x + 1}}{{(x + 1)(x - 2) + 4}} \hfill \\ \end{gathered} $$

And for $x \geqslant 3$ we get for RHS:

$$\frac{1}{4} \cdot \frac{{x + 1}}{{(x + 1)(x - 2) + 4}} < \frac{1}{4} \cdot \frac{{x + 1}}{{(x + 1)(x - 2)}} = \frac{1}{4} \cdot \frac{1}{{x - 2}}$$

It remains to show $$\mathop {\lim }\limits_{x \to \infty } \frac{1}{{x - 2}} = 0$$