Daily Math · 2026-06-02

Question 1

For $f(x) = \begin{cases} x + 1, & x < 0 \ x^2 + 1, & x \ge 0 \end{cases}$, $f$ is continuous at $x = 0$ because:

Accepted Answer

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

Answer

$f(0)$ is undefined

Answer

The left and right limits differ

Answer

$\lim_{x 	o 0^-}f = \lim_{x 	o 0^+}f = f(0) = 1$

Answer

$f$ is a polynomial

Question 2

The solution set of $|x - 2| < 3$ is:

Accepted Answer

$|x - 2| < 3 \Leftrightarrow -3 < x - 2 < 3 \Leftrightarrow -1 < x < 5$. The solution set is $(-1, 5)$.

Answer

$(-3, 3)$

Answer

$(-1, 5)$

Answer

$(2, 5)$

Answer

$[-1, 5]$

Question 3

Evaluate $\displaystyle\lim_{x	o 0} \dfrac{1 - \cos(x)}{x^2}$.

Accepted Answer

Using $1-\cos x = 2\sin^2(x/2)$: $\displaystyle\lim_{x	o 0}\dfrac{2\sin^2(x/2)}{x^2} = \dfrac{1}{2}\lim_{x	o 0}\left(\dfrac{\sin(x/2)}{x/2}ight)^2 = \dfrac{1}{2}\cdot 1 = \dfrac{1}{2}$.

Answer

$0$

Answer

$\dfrac{1}{4}$

Answer

$\dfrac{1}{2}$

Answer

$1$

Problems & worked solutions

Problem 1 — Continuity

Problem 2 — Linear Function

Problem 3 — Notable Limits