Easy L'Hopitals question

[faust9]

Howdy, I'm stumped by a seemingly easy L'Hopitals limit question.

\lim_{t\rightarrow 0}\ \frac{\ln t}{t^2-1}

I said the limit doesn't exist, but the asker claims it does. I tried various transformations so that I could use L'Hopitals theorm but with no success. I keep getting \frac{- \infty}{-1} so I can't take the derivative of the top and the bottom to get the answer.

Any help would be greatly appreciated.

[mathman]

I think that the problem statement has a minor(?) error, it should be limit for t->1, not 0. Then L'Hospital's can be used (to get 1/2).

[faust9]

That's what I thought too, but the question was worded as I stated it...

Anyway, thanks for the input.

[HallsofIvy]

As stated, L'Hopital's rule does not apply: the numerator goes to infinity and the denominator goes to 1 so the limit does not exist.

The limits as x->1 or as x-> infinity both exist and can be done by L'Hopital. As mathman said, the limit at 1 is 1/2. The limit at
infinity is 0.

[Spectre5]

(-infinity)/(-1) is not an indeterminate form....it equals positive infinity

so the limit can be computed by simply plugging zero and then getting positive infintity