Daily Math · 2026-06-17

Question 1

The vertical asymptote of $f(x) = \dfrac{1}{x - 3}$ is the line:

Accepted Answer

$x - 3 = 0 \Rightarrow x = 3$.

Answer

$y = 0$

Answer

$y = 3$

Answer

$x = 3$

Answer

$x = 0$

Question 2

$\displaystyle\lim_{x 	o 0} \dfrac{\sin 3x}{	an 5x}$ equals:

Accepted Answer

$\dfrac{\sin 3x}{	an 5x} = \dfrac{\sin 3x}{3x} \cdot \dfrac{5x}{	an 5x} \cdot \dfrac{3}{5} 	o 1 \cdot 1 \cdot \dfrac{3}{5} = \dfrac{3}{5}$.

Answer

$	frac{3}{5}$

Answer

$	frac{5}{3}$

Answer

$1$

Answer

$0$

Question 3

Let $f: \mathbb{R} 	o \mathbb{R}$ be defined by $f(x) = \begin{cases} x + 1, & x < 0 \ 2x + 1, & x \ge 0 \end{cases}$. Which statement is correct?

Accepted Answer

**Left branch** ($x < 0$): $f(x) = x + 1 \in (-\infty,\, 1)$, strictly increasing.
**Right branch** ($x \ge 0$): $f(x) = 2x + 1 \in [1,\, +\infty)$, strictly increasing.

The two images $(-\infty, 1)$ and $[1, +\infty)$ are disjoint and together cover all of $\mathbb{R}$, so $f$ is **surjective**. Each branch is strictly increasing, and the ranges don't overlap, so no two distinct $x$-values give the same $f(x)$ — $f$ is **injective**. Hence $f$ is bijective.

(Note: $f(-1) = 0$ and $f(0) = 1$, so distractor A is factually false.)

Answer

$f$ is not injective because $f(-1) = f(0)$

Answer

$f$ is injective but not surjective

Answer

$f$ is surjective but not injective

Answer

$f$ is bijective

Problems & worked solutions

Problem 1 — Asymptotes

Problem 2 — Calculus

Problem 3 — Functions — General Properties