## Calculus: Early Transcendentals 8th Edition

If $~~\vert x+3 \vert \lt \frac{1}{2}~~$ then $~~\vert 4x+13 \vert \lt 3$
Suppose that $\vert x+3 \vert \lt \frac{1}{2}$ Then: $\vert x+3 \vert \lt \frac{1}{2}$ $-\frac{1}{2} \lt x+3 \lt \frac{1}{2}$ $-2 \lt 4x+12 \lt 2$ $-1 \lt 4x+13 \lt 3$ $-3 \lt -1 \lt 4x+13 \lt 3$ $\vert 4x+13 \vert \lt 3$