Question

How do you find the limit of `(8x -lnx)` as x approaches infinity?

Answer 1

Use `8x-lnx = x(8-lnx/x)`.

Explanation:

`8x-lnx = x(8-lnx/x)`.

`lim_(xrarroo)lnx/x=0` `" "` (Use l"Hospital if you don't know this limit.)

So

`lim_(xrarroo)(8x-lnx) = lim_(xrarroo)(x(8-lnx/x)) = oo(8-0) = oo`.

I'm not that comfortable using this notation, but it's so convenient!

Here it is expressed without the notation.

As `x` increases without bound, `8-lnx/x rarr 8`,

so `x(8-lnx/x)` is the product of an expression whose value increases without bound and an expression whose value approaches a positive value. The product, therefore increases without bound.