Question

evaluate the limit.
lim_{t \to -infty} left(9 t^{2}+8 t+5
ight)
simplify any fractions in your answer.

Explanation:

Step1: Analyze the leading term

The leading term of \(9t^{2}+8t + 5\) is \(9t^{2}\). As \(t\to-\infty\), we consider the behavior of \(y = 9t^{2}\).
For \(y = 9t^{2}\), when \(t\to-\infty\), \(t^{2}=(-t)\times(-t)\to+\infty\). And since the coefficient \(a = 9>0\), \(9t^{2}\to+\infty\).
The terms \(8t\) and \(5\) are negligible compared to \(9t^{2}\) as \(t\to-\infty\). Mathematically, we can use the property \(\lim_{t\to-\infty}(a_{n}t^{n}+a_{n - 1}t^{n-1}+\cdots+a_{0})=\lim_{t\to-\infty}a_{n}t^{n}\) for \(n>0\) and \(a_{n}
eq0\).

Answer:

\(+\infty\)