Given that $-4\leq x\leq-2$ and $2\leq y\leq4$, what is the largest possible value of $\frac{x+y}{x}$?
$\mathrm{(A) \ } -1 \qquad \mathrm{(B) \ } -\frac12 \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac12 \qquad \mathrm{(E) \ } 1$
Rewrite $\frac{(x+y)}x$ as $\frac{x}x+\frac{y}x=1+\frac{y}x$.
We also know that $\frac{y}x<0$ because $x$ and $y$ are of opposite sign.
Therefore, $1+\frac{y}x$ is maximized when $|\frac{y}x|$ is minimized, which occurs when $|x|$ is the largest and $|y|$ is the smallest.
This occurs at $(-4,2)$, so $\frac{x+y}x=1-\frac12=\frac12\Rightarrow\boxed{\mathrm{(D)}\ \frac{1}{2}}$.