So I understand that just because a function could be undefined does not mean that the limit is undefined, but how would you go about solving this:

Evaluate the limit, if it exists:

limit as t goes to 0 (1/t - 1/t^2+t)


And if it does not exist(which I understand that it still might exist) how would I prove it?

compute the limit from the left and the limit from the right.
if they're equal and it's not undefined, then the limit exists. otherwise, it doesn't exist.

quick and dirty trick.

for the limit as x goes to L of f(x), compute f(x + epsilon) and f(x - epsilon). if it's clear it approaches the same value, then it probably has a limit. if they're radically different or they're approaching infinity/negative infinity, then a limit doesn't exist.

epsilon is a small value. i'd pick 0.000000001 or something like that.

This post was edited by carteblanche on Sep 9 2015 06:50pm

Look at each term seperately

1 / t as t goes to 0 would be infinity

-1 / t^2 as t goes to 0 would be negative infinity, and it would go negative much faster then the above term because the t is squared

t as t goes to 0 is 0

So you have something going to infinty + something going to negative infinity way faster + 0 = Negative infinity

This post was edited by RzChaos on Sep 9 2015 06:42pm

To have it proven, transform your expression :

1/t - 1/t² + t = ( t - 1 + t^3 ) / t²

Now it's clear that the numerator of the fraction tends to - 1 as t approaches zero, while the denominator tends to "positive" zero.
By usual rules, all this tends to negative infinity, as t approaches zero by any side (right or left).
Right and left limits being equal, it's safe to say that the limit exists, and is - ∞.