Daily math problems for June 17, 2026 — Calculus, Limits, Functions

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

${\displaystyle \frac{\sin 3x}{\tan 5x}}={\displaystyle \frac{\sin 3x}{3x}}\cdot {\displaystyle \frac{5x}{\tan 5x}}\cdot {\displaystyle \frac{3}{5}}\to 1\cdot 1\cdot {\displaystyle \frac{3}{5}}={\displaystyle \frac{3}{5}}$.

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

The two images $(-\mathrm{\infty },1)$ and $[1,+\mathrm{\infty })$ 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.)