Take f[x] = (3 x - 1)/Sqrt[x^2 + x +1]. Find a good global scale picture of the graph of f[x] and describe what you see.

Answer:

Clear[f];
f[x_]=(3 x - 1)/Sqrt[x^2 + x +1]

Now plot f[x] over several, increasingly larger ranges for x values:

Plot[f[x],{x,-100,100}, PlotRange->All]

Plot[f[x],{x,-1000,1000}, PlotRange->All]

This example shows that we can get different horizontal asymptotes for large positive values of x than we get for large negative values of x. The graph clearly indicates that
f[x] = 3 and f[x] = -3.