This time it is not so obvious what the limit value is. We could substitute larger and larger values of x until we could see what was happening (try `100`, then `1\ 000`, then `1\ 000\ 000` and so on).

Or, we could rearrange the expression and use the fact that

`lim_(x->oo)(1/x)=0`

to find the limiting value.

We divide throughout by x to get the expression in a form where we can evaluate it.

`lim_(x->oo)((5-3x)/(6x+1))`

`=lim_(x->oo)((5/x-3)/(6+1/x))`

`=(0-3)/(6+0)`

`=-1/2`

Notice that we cannot substitute ∞ into the fraction `((5/x-3)/(6+1))`, because this does not make mathematical sense.

Please do not write `((5-3(oo))/(6(oo)+1))`. It really upsets mathematicians.