Question

Find the limit:

Answer 1

\[\lim_{t \rightarrow 0}\frac{ t^2-+t-2 }{ (t-1) }\]

Answer 2

The bottom is an absolute value of (t-1). Wasn't sure how to do the absolute value sign in the equation format.

Answer 3

oops, the top is \[t^2+t-2\]

Answer 4

just stuff in t = 0, see where you go.

and sort out the denominator:-) you can say |x| = absolute value of x.

Answer 5

\[\lim_{t \rightarrow 0}\frac{ t^2+t-2 }{ \left| t-1 \right| }\]
\[when~ t \rightarrow 0,t-1<0,\]
\[\lim_{t \rightarrow 0}\frac{ t^2+t-2 }{ -t+1 }=\frac{ 0+0-2 }{ -0+1 }=-2\]

Answer 6

\[t -> 0, t-1 < 0,\] How come it is -t+1 instead of -(t+1) = -t - 1? Other then that, I understand the problem.

Answer 7

if \[\left| x-a \right|<0,`then~\left| x=a \right|=-\left( x-a \right)=-x+a\]

Answer 8

Oh, I made a slight error with the "+" and "-," I read it wrong. I understand completely now. THanks! :)

Answer 9

correction
\[\left| x-a \right|=-\left( x-a \right)=-x+a\]

Answer 10

One more question, would we have to solve for t-1 > 0 next or just t-1 < 0?

Answer 11

@surjithayer

Answer 12

as t approaches 0
0<1
hence t<1
t-1<0
so in both ways t-1<0
i.e.,t approaches 0-and t approaches 0+

Answer 13

@surjithayer
To clarify, there are two answers: -2 & 2?

Answer 14

you meant to ask this, i think:

\[\lim_{t \rightarrow 0}\frac{ t^2-+t-2 }{ |t-1| }\]?

so if t = 0 ??????

you have what?