Question

evaluate the limit using the appropriate limit law(s). (if an answer does not exist, enter dne.)
lim_{t
ightarrow2}\frac{t^{4}-2}{2t^{2}-3t + 1}

Explanation:

Step1: Substitute $t = 2$ into the function

First, substitute $t = 2$ into the numerator $t^{4}-2$ and the denominator $2t^{2}-3t + 1$.
For the numerator: $2^{4}-2=16 - 2=14$.
For the denominator: $2\times2^{2}-3\times2 + 1=2\times4-6 + 1=8 - 6+1=3$.

Step2: Calculate the limit value

Since substituting $t = 2$ into the rational - function $\frac{t^{4}-2}{2t^{2}-3t + 1}$ gives a well - defined real number, the limit is equal to the value of the function at $t = 2$.
The limit $\lim_{t
ightarrow2}\frac{t^{4}-2}{2t^{2}-3t + 1}=\frac{14}{3}$.

Answer:

$\frac{14}{3}$