Question

use lhôpitals rule to find the limit.
lim_{t \to -5} \frac{t^{3}-2 t + 115}{t^{2}-t - 30}
lim_{t \to -5} \frac{t^{3}-2 t + 115}{t^{2}-t - 30}=quad \text { (type an exact answer.) }

Explanation:

Step1: Check the form of the limit

Substitute \(t = - 5\) into \(\frac{t^{3}-2t + 115}{t^{2}-t - 30}\).
For the numerator: \((-5)^{3}-2\times(-5)+115=-125 + 10+115=0\).
For the denominator: \((-5)^{2}-(-5)-30=25 + 5-30=0\).
Since we get the \(\frac{0}{0}\) indeterminate form, we can apply L'Hôpital's Rule.

Step2: Differentiate the numerator and denominator

Differentiate the numerator \(y_1=t^{3}-2t + 115\) using the power rule \((x^n)^\prime=nx^{n - 1}\).
\(y_1^\prime=\frac{d}{dt}(t^{3}-2t + 115)=3t^{2}-2\).
Differentiate the denominator \(y_2=t^{2}-t - 30\) using the power rule.
\(y_2^\prime=\frac{d}{dt}(t^{2}-t - 30)=2t-1\).

Step3: Find the new limit

Now, find \(\lim_{t
ightarrow - 5}\frac{3t^{2}-2}{2t - 1}\).
Substitute \(t=-5\) into \(\frac{3t^{2}-2}{2t - 1}\).
\(\frac{3\times(-5)^{2}-2}{2\times(-5)-1}=\frac{3\times25-2}{-10 - 1}=\frac{75-2}{-11}=\frac{73}{-11}=-\frac{73}{11}\).

Answer:

\(-\frac{73}{11}\)