Question

How do you find the limit of `lnx/(sqrtx+lnx)` as `x->oo`?

Answer 1

`lim_(xrarroo)lnx/(sqrtx+lnx)=0`

Explanation:

`L=lim_(xrarroo)lnx/(sqrtx+lnx)`

This is an indeterminate form, in the form of `oo/oo`. Due to that, we can use L'Hopital's rule by differentiating the numerator and denominator individually.

`L=lim_(xrarroo)(d/dxlnx)/(d/dx(sqrtx+lnx))=lim_(xrarroo)(1/x)/(1/(2sqrtx)+1/x)`

Multiply through by `x/x`:

`L=lim_(xrarroo)(1/x*x)/((1/(2sqrtx)+1/x)*x)=lim_(xrarroo)1/(sqrtx/2+1)`

As we examine this, we see the denominator approaches infinity as the numerator stays unchanged, so the overall fraction will approach `0`.

`L=0`

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Another way of approaching this is that both the numerator and denominator of the fraction contain `lnx`, so those will increase at the same rate. However, the denominator is also increasing with `sqrtx`, so the denominator will be growing faster than the numerator thanks to the additional `sqrtx`. Thus, the limit will be `0` because the denominator will constantly be outgrowing the numerator.