Given $\|\mathbf{v}\| = 5$ and $\|\mathbf{w}\| = 8,$ find the largest possible value of
\[\|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\|.\]
Note that
\begin{align*}
\operatorname{proj}_{\mathbf{w}} \mathbf{v} &= \left\| \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{w}\|^2} \mathbf{w} \right\| \\
&= \frac{|\mathbf{v} \cdot \mathbf{w}|}{\|\mathbf{w}\|^2} \cdot \|\mathbf{w}\| \\
&= \frac{|\mathbf{v} \cdot \mathbf{w}|}{\|\mathbf{w}\|}.
\end{align*}Let $\theta$ be the angle between $\mathbf{v}$ and $\mathbf{w}.$  Then $\mathbf{v} \cdot \mathbf{w} = \|\mathbf{v}\| \|\mathbf{w}\| \cos \theta,$ so
\[\frac{|\mathbf{v} \cdot \mathbf{w}|}{\|\mathbf{w}\|} = \frac{|\|\mathbf{v}\| \|\mathbf{w}\| \cos \theta|}{\|\mathbf{w}\|} = \|\mathbf{v}\| |\cos \theta| = 5 |\cos \theta| \le 5.\]Equality occurs when $\theta = 0,$ or when $\mathbf{v}$ and $\mathbf{w}$ point in the same direction, so the largest possible value is $\boxed{5}.$

If $\mathbf{p}$ is the projection of $\mathbf{v}$ onto $\mathbf{w},$ then the length of $\mathbf{p}$ is the leg of a right triangle that has the length of $\mathbf{v}$ as the hypotenuse.  Thus, geometrically, $\|\mathbf{p}\| \le \|\mathbf{v}\| = 5.$

[asy]
unitsize(1 cm);

pair O, P, V, W;

O = (0,0);
V = (10/7,sqrt(3^2 - (10/7)^2));
P = (10/7,0);
W = (7,0);

draw(O--V,Arrow(6));
draw(O--W,Arrow(6));
draw(O--P,red,Arrow(6));
draw(P--V,dashed);

label("$\mathbf{v}$", V, N);
label("$\mathbf{w}$", W, E);
label("$\mathbf{p}$", P, S);
[/asy]