How do you find the limit of #(t^2+t-2)/(t^2-1)# as #t-1#?

Question

How do you find the limit of #(t^2+t-2)/(t^2-1)# as #t->1#?

Ezra Adams · Accepted Answer

#3/2# factorization initially #((t+2)(t-1))/((t-1)(t+1))#. # (t-1)#gets cancelled out. Thus it is, #lim_(t->1) (t+2)/(t+1)# =#3/2#

Emilia Aguilar · Answer

undefined To find the limit of (t^2+t-2)/(t^2-1) as t approaches 1, we can substitute the value of 1 into the expression. However, this would result in division by zero, which is undefined. Therefore, we need to simplify the expression before substituting t=1. Factoring the numerator and denominator, we get (t+2)(t-1)/(t+1)(t-1). Canceling out the common factor of (t-1), we are left with (t+2)/(t+1). Now, substituting t=1 into this simplified expression, we find that the limit is equal to 3/2.

HIX Tutor · Answer

undefined That's a great question! Let’s break it down step-by-step.

1. **Substitute t = 1** in the function:  
   $$
   \frac{(1^2 + 1 - 2)}{(1^2 - 1)} = \frac{(1 + 1 - 2)}{(1 - 1)} = \frac{0}{0}
   $$
   This means we get an "undefined" form. So, we need to simplify.

2. **Factor the numerator and the denominator**:  
   The numerator $t^2 + t - 2$ can be factored.  
   What two numbers multiply to -2 and add to +1? The numbers are +2 and -1.

So, we can write it as:  
   $$
   (t + 2)(t - 1)
   $$

The denominator $t^2 - 1$ is a difference of squares:  
   It factors to:  
   $$
   (t - 1)(t + 1)
   $$

3. **Rewrite the expression**:  
   Now, we have:  
   $$
   \frac{(t + 2)(t - 1)}{(t - 1)(t + 1)}
   $$

4. **Cancel the common factor**:  
   We can cancel $t - 1$ from the numerator and denominator (as long as $t \neq 1$):  
   $$
   \frac{t + 2}{t + 1}
   $$

5. **Find the limit as t approaches 1**:  
   Now substitute $t = 1$ into the simplified expression:  
   $$
   \frac{1 + 2}{1 + 1} = \frac{3}{2}
   $$

6. **Final answer**:  
   So, the limit of $\frac{t^2 + t - 2}{t^2 - 1}$ as $t$ approaches 1 is:  
   $$
   \frac{3}{2}
   $$

Does that make sense?

How do you find the limit of #(t^2+t-2)/(t^2-1)# as #t-&gt;1#?

How do you find the limit of #(t^2+t-2)/(t^2-1)# as #t->1#?