Questions: Evaluate the limit [ lim t rightarrow 2 fract^2-42 t^2-7 t+6 ] Enter DNE if the limit does not exist.

$Evaluate the limit [ lim t rightarrow 2 fract^2-42 t^2-7 t+6 ] Enter DNE if the limit does not exist.$

Transcript text: Assignment1W24: Problem 8 (1 point) Evaluate the limit \[ \lim _{t \rightarrow 2} \frac{t^{2}-4}{2 t^{2}-7 t+6} \] Enter DNE if the limit does not exist.

Solution

Solution Steps

Step 1: Define the Limit

We need to evaluate the limit: \[ \lim_{t \rightarrow 2} \frac{t^{2}-4}{2t^{2}-7t+6} \]

Step 2: Factor the Numerator and Denominator

The numerator $ t^{2} - 4 $ can be factored as: \[ t^{2} - 4 = (t - 2)(t + 2) \] The denominator $ 2t^{2} - 7t + 6 $ can be factored as: \[ 2t^{2} - 7t + 6 = (2t - 3)(t - 2) \]

Step 3: Simplify the Expression

Substituting the factored forms into the limit gives: \[ \lim_{t \rightarrow 2} \frac{(t - 2)(t + 2)}{(2t - 3)(t - 2)} \] We can cancel the common factor $ (t - 2) $ (noting that $ t \neq 2 $): \[ \lim_{t \rightarrow 2} \frac{t + 2}{2t - 3} \]

Step 4: Substitute $ t = 2 $

Now we substitute $ t = 2 $ into the simplified expression: \[ \frac{2 + 2}{2(2) - 3} = \frac{4}{4 - 3} = \frac{4}{1} = 4 \]

Step 5: Conclusion

Thus, the limit is: \[ \lim_{t \rightarrow 2} \frac{t^{2}-4}{2t^{2}-7t+6} = 4 \]