Question

\[\lim x \rightarrow1 t+1 /(t-1)^2\]

Answer 1

\[\lim_{t \rightarrow 1}{t+1 \over (t-1)^2}=\infty\]

Answer 2

yep! How'd you get it?

Answer 3

Plugging \(t=1\) gives \(\frac{2}{0}\), which in undefined. Whenever you have a number other than \(0\),over \(0\) then the limit goes to \(\infty\) or\( - \infty\). All you have to do is to test the sign around \(1\), and you can see it's positive as x approaches 1 from both sides.

Answer 4

I thought I'd have to factor it and get a real non-zero answer at first.

Answer 5

when do you factor and try to get a non-zero denominator then?

Answer 6

what?

Answer 7

differentiate top and bottom :|

Answer 8

you can't differentiate top and bottom

Answer 9

its in the limits section, so not up to that yet.

Answer 10

you can't apply l'hopital because it's not 0/0

Answer 11

ohh yeh, twice and its zero lol

Answer 12

You should try factoring when you have a \(\frac{0}{0}\) or a \(\frac{\infty}{0}\) or a similar form.

Answer 13

oh -- thank you. I've been factoring whenever I get 0 in the denominator.