What do when you got this?

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I assume you are trying to find the limit of a function, and then you get Ln (infinity/infinity). You just do it in the same way you would do in infinity/infinity, cuz Limit (ln(f(x)))=ln (Limit (f (x))) for fx continous functions.

>>140628
But the natural logarithm isn't continuous in x=0.

>>140633
Is f (0)->0 when x->0? Cuz its
Ln (f(x)) not ln (x). Is this a highschool level problem? Ignore the continous if yes. If x->0 then ln (x)->(-infinity).

The functions are:
lim x-> infinity from ln((x-1)/x^2)

and

lim x-> - infinity from ln((x-1)/x^2)

>>140628
That means that I should use L' Hospital and solve this?
Ok, thanks for help. I completely forgot that rule.

The answer to the first one is -infinity
. (x-1)/x^2->0 when x->0. So lim(ln(f(x)) )when x->infinity is the same as lim ln (x) when x->0 which is -infinity. The second one doesn't make sense. The function is not defined for x <1.