Daily math problems for June 2, 2026 — Calculus, Linear Function, Limits

${\lim }_{x\to 0^{-}}(x+1)=1$, ${\lim }_{x\to 0^{+}}(x^2+1)=1$, $f(0)=1$. All three agree, so $f$ is continuous at $0$.

$\mathrm{\mid }x-2\mathrm{\mid }<3\iff -3<x-2<3\iff -1<x<5$. The solution set is $(-1,5)$.

Using $1-\cos x=2{\sin }^2(x\mathrm{/}2)$: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\underset{x\to 0}{\lim }{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2={\displaystyle \frac{1}{2}}\cdot 1={\displaystyle \frac{1}{2}}}$.