Solving Easy Limit Problem: \mathop{\lim}\limits_{t\to 1} \frac{t-1}{t^2-1}

[tony873004]

\mathop {\lim }\limits_{t \to 1} \,\frac{{t - 1}}{{t^2 - 1}}

I thought we were taught to simply divide the coefficients of the highest term, in this case, 0t² for the numerator and 1 t² for the denominator. 0/1=0. But I know the limit is 0.5 from substituting 0.9999999999 for t in my calculator.

I must be getting this "coefficient of highest term" method mixed up with something else. Why doesn't it work here?

 

[sutupidmath]

well what u need to do is
t^{2}-1=(t-1)(t+1)
can u go from here?
This is just the difference of squares. Its general form is:

a^{2}-b^{2}=(a-b)(a+b)

 

[HallsofIvy]

\mathop {\lim }\limits_{t \to 1} \,\frac{{t - 1}}{{t^2 - 1}}

I thought we were taught to simply divide the coefficients of the highest term, in this case, 0t² for the numerator and 1 t² for the denominator. 0/1=0. But I know the limit is 0.5 from substituting 0.9999999999 for t in my calculator.

I must be getting this "coefficient of highest term" method mixed up with something else. Why doesn't it work here?

I strongly suspect you were taught that for limits as t goes to infinity! That is not the problem here.

 

[tony873004]

Thanks for the explanation, sutupid.

You're probably right, Halls. I'll never forget this now. This was part of a larger problem in Calc III. That's the problem with Calc III. Every now and then they assume you remember your Calc I :)

 

[sutupidmath]

Every now and then they assume you remember your Calc I :)

Honestly, this had almost nothing to do with calc I, it was just a simple algebra trick!

 

[HallsofIvy]

Well, the tiny part about finding the limit might be from calculus I!
(And you did say "almost nothing".)

 

[tony873004]

Algebra class was 20 years ago for me. Everything I currently know about Algebra I learned in Calc I.