Use the \(\varepsilon\)-\(\delta\) definition of the limit to prove that \[ \lim_{x\to1}\frac{x^2+1}{x}=2. \]
Solution
Let \(\varepsilon>0\) and choose \[ \delta=\min\left\{\frac12,\sqrt{\frac\varepsilon2}\right\}. \] If \(|x-1|<\delta\), then \(|x|>1/2\). Hence \[ \left|\frac{x^2+1}{x}-2\right| =\frac{|x-1|^2}{|x|} <2|x-1|^2 <2\delta^2 \leq\varepsilon. \]