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anonymous
 5 years ago
\[\lim x \rightarrow1 t+1 /(t1)^2\]
anonymous
 5 years ago
\[\lim x \rightarrow1 t+1 /(t1)^2\]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{t \rightarrow 1}{t+1 \over (t1)^2}=\infty\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep! How'd you get it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Plugging \(t=1\) 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought I'd have to factor it and get a real nonzero answer at first.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when do you factor and try to get a nonzero denominator then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate top and bottom :

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can't differentiate top and bottom

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its in the limits section, so not up to that yet.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can't apply l'hopital because it's not 0/0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh yeh, twice and its zero lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You should try factoring when you have a \(\frac{0}{0}\) or a \(\frac{\infty}{0}\) or a similar form.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh  thank you. I've been factoring whenever I get 0 in the denominator.
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