Question

How do you find the Limit of `(x^2)/(lnx)` as x approaches infinity?

Answer 1

It is `oo`.

Explanation:

These two functions goes to infinity when `x->oo` so it is in the form `oo/oo`.
Then we can appply the rule of L'Hôpital

`lim_(x->oo)(x^2)/ln(x)=lim_(x->oo)(d/dxx^2)/(d/dxln(x))`
`=lim_(x->oo)(2x)/(1/x)=lim_(x->oo)2x^2=oo`

Then we discovered that `x^2` goes to infinity "faster" than `ln(x)` and the ratio goes to infinity.

Answer 2

I got `lim_(x->oo) (x^n)/(lnx) = lim_(x->oo) nx^n = oo`.

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Note that

`lim_(x->oo) (x^2)/(lnx)`

is of the form `oo/oo`, since plugging in arbitrarily large numbers into `x^2` and `lnx` gives a large number as well.

This fits the bill for when we can use L'Hopital's Rule (for `oo/oo` and `0/0` only!).

This rule states:

Taking the individual derivative of the numerator and denominator retains the same limit as if you didn't, if and only if the expression is of the form `oo/oo` or `0/0` and both functions are continuous in the relevant closed interval.

So, in, say, `[2,oo)`:

`color(green)(lim_(x->oo) (x^n)/(lnx))`

`= lim_(x->oo) (nx^(n-1))/(1/x)`

`= lim_(x->oo) nx^(n-1)*(1/x)^(-1)`

`= lim_(x->oo) nx^(n-1)*x^1`

`= color(green)(lim_(x->oo) nx^n)`

This turned out this way because `x^n` accelerates towards `oo`, whereas `lnx` tapers off as it approaches `oo`. Hence, at arbitrarily large numbers, `\mathbf(x^n)` dominates.

This means for `n = 2`, we still have:

graph{x^2/(lnx) [-5, 25, -142, 182.6]}

`color(blue)(lim_(x->oo) (x^2)/(lnx))`

`= lim_(x->oo) (2x)/(1/x)`

`= lim_(x->oo) 2x^2`

`= 2 lim_(x->oo) x^2`

`= 2oo^2`

`=> color(blue)(oo)`

And Wolfram Alpha agrees with me.