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

• tony873004
In summary, the limit as t approaches 1 of (t-1)/(t^2-1) is 0.5. This is because t^2-1 can be factored into (t-1)(t+1), and the limit can then be evaluated by substituting 1 for t. The "coefficient of highest term" method is not applicable here because it is meant for limits as t approaches infinity. It is important to remember algebra tricks, even if they were learned in a previous class, as they can be useful in more advanced courses.
tony873004
Gold Member
$$\mathop {\lim }\limits_{t \to 1} \,\frac{{t - 1}}{{t^2 - 1}}$$

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?

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:

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

Last edited:
tony873004 said:
$$\mathop {\lim }\limits_{t \to 1} \,\frac{{t - 1}}{{t^2 - 1}}$$

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.

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 :)

tony873004 said:
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!

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

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

## What is the definition of a limit?

A limit is a mathematical concept that describes the behavior of a function as its input approaches a specific value or point. In other words, it represents the value that a function is approaching as its input gets closer and closer to a certain point.

## How do you solve limit problems?

To solve a limit problem, you can use algebraic manipulation, graphing, or substitution to evaluate the function at the given point. If the function is undefined at that point, you can use the limit laws or L'Hopital's rule to find the limit.

## What is the difference between left and right limits?

A left limit represents the value that a function is approaching from the left side of a given point, while a right limit represents the value that a function is approaching from the right side of a given point. In order for the limit to exist, the left and right limits must be equal.

## What is the Squeeze Theorem?

The Squeeze Theorem, also known as the Sandwich Theorem, is a mathematical theorem that states if a function is "squeezed" between two other functions that have the same limit at a given point, then the squeezed function must also have the same limit at that point.

## How do you solve the limit problem: \mathop{\lim}\limits_{t\to 1} \frac{t-1}{t^2-1}?

To solve this limit problem, we can first factor the numerator and denominator of the fraction to get \frac{(t-1)}{(t-1)(t+1)}. We can then cancel out the common factor of (t-1) to get \frac{1}{t+1}. Finally, we can substitute the given value of t=1 into the simplified expression to get \frac{1}{1+1} = \frac{1}{2}. Therefore, the limit as t approaches 1 of \frac{t-1}{t^2-1} is equal to \frac{1}{2}.

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