Question

What is the limits?

Indeterminate Forms

Answer 1

Use either L'hopital's rule, or apply our understanding of the rate of increase of `ln(x)` relative to the increase in `x`. The limit is 0.

Explanation:

There are two ways of doing this.

We will use L'hopital's rule, which states that for `lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))`.

Then...

`d/dx (x ln x) = d/dx (x)*lnx + x *d/dx (lnx) = 1*lnx + x/x = lnx +1`

`d/dx (x^2+1) = 2x...`

`(f'(x))/(g'(x)) = (lnx +1)/(2x)`

This is still not conclusive, so we must perform another iteration of L'hopital's rule...

`d/dx (ln(x)+1) = 1/x, d/dx 2x = 2`

Then `(f''(x))/(g''(x)) = (1/x)/2 = 1/(2x)`

The limit of this function, as easily observed, will be `lim_(x->oo) 1/(2x) = 0`

The second method requires less work, but also requires a stronger understanding. We know that `ln(x) < x` for all x, and that `ln x` increases very slowly by comparison to `x`. We could, depending on how careful we wanted to be, essentially ignore ln x in the numerator for this purpose; the main determining factor in the magnitude of the numerator as `x->oo` is `x`. We can also ignore the +1 on the bottom, as it will rapidly become negligible. Then, if we ignore `lnx`...

`lim_(x->oo) (xlnx)/(x^2+1) approx lim_(x->oo)(x)/(x^2) = lim_(x->oo) 1/x = 0`