Suppose \(f: \mathbb{R} \to \mathbb{R}\).
- Give the \(\delta\)-\(\varepsilon\) definition of \(\lim_{x \to a} f(x) = \ell\).
- Determine the limit and use the definition to prove your answer: \[\lim_{x \to 2} \frac{2x+1}{x^2+1} = ?\]
Solution
The statement \(\lim_{x\to a}f(x)=\ell\) means: for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ 0<|x-a|<\delta\implies |f(x)-\ell|<\varepsilon. \]
Here the candidate limit is \[ \frac{2(2)+1}{2^2+1}=1. \] Moreover \[ \left|\frac{2x+1}{x^2+1}-1\right| =\frac{|x(2-x)|}{x^2+1}. \] If \(|x-2|<1\), then \(|x|<3\), while \(x^2+1\ge1\). Thus \[ \left|\frac{2x+1}{x^2+1}-1\right|\le3|x-2|. \] Given \(\varepsilon>0\), choose \[ \delta=\min\{1,\varepsilon/3\}. \] Then \(0<|x-2|<\delta\) implies the displayed error is \(<\varepsilon\). Hence the limit is \(1\).