Question

use lhôpitals rule to find the limit.

$$\lim_{t\to - 5}\frac{t^{3}-6t + 95}{t^{2}-t - 30}$$

$$\lim_{t\to - 5}\frac{t^{3}-6t + 95}{t^{2}-t - 30}=\square$$ (type an exact answer.)

Explanation:

Step1: Check the form of the limit

Substitute \(t = - 5\) into \(\frac{t^{3}-6t + 95}{t^{2}-t - 30}\).
For the numerator: \((-5)^{3}-6\times(-5)+95=-125 + 30+95=0\).
For the denominator: \((-5)^{2}-(-5)-30=25 + 5-30=0\).
We have the \(\frac{0}{0}\) indeterminate form, so we can apply L'Hôpital's Rule.

Step2: Differentiate the numerator and denominator

Differentiate the numerator \(y=t^{3}-6t + 95\) using the power rule \((x^{n})^\prime=nx^{n - 1}\). The derivative \(y^\prime=3t^{2}-6\).
Differentiate the denominator \(y=t^{2}-t - 30\). The derivative \(y^\prime=2t-1\).

Step3: Find the new limit

Now we find \(\lim_{t
ightarrow - 5}\frac{3t^{2}-6}{2t - 1}\).
Substitute \(t=-5\) into \(\frac{3t^{2}-6}{2t - 1}\):

$$
LATEXBLOCK0
$$

Answer:

\(-\frac{69}{11}\)