Question

use lhôpitals rule to find the following limit.
lim(h→0) (e^(3h)-(1 + 3h))/(5h^2)
lim(h→0) (e^(3h)-(1 + 3h))/(5h^2)= (type an exact answer.)

Explanation:

Step1: Check indeterminate form

When \(h
ightarrow0\), the numerator \(e^{3h}-(1 + 3h)=e^{0}-(1+0)=1 - 1=0\) and the denominator \(5h^{2}=0\). So, it is in the \(\frac{0}{0}\) indeterminate - form and L'Hôpital's Rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of the numerator \(y_1 = e^{3h}-(1 + 3h)\) using the chain - rule. The derivative of \(e^{3h}\) is \(3e^{3h}\) and the derivative of \((1 + 3h)\) is 3. So, \(y_1^\prime=3e^{3h}-3\). The derivative of the denominator \(y_2 = 5h^{2}\) is \(y_2^\prime = 10h\). Now, the limit becomes \(\lim_{h
ightarrow0}\frac{3e^{3h}-3}{10h}\).

Step3: Check indeterminate form again

When \(h
ightarrow0\), \(3e^{3h}-3=3e^{0}-3=3 - 3 = 0\) and \(10h = 0\). So, it is still in the \(\frac{0}{0}\) indeterminate - form, and we apply L'Hôpital's Rule again.

Step4: Differentiate numerator and denominator again

The derivative of the numerator \(y_3=3e^{3h}-3\) is \(y_3^\prime = 9e^{3h}\), and the derivative of the denominator \(y_4 = 10h\) is \(y_4^\prime=10\). Now, the limit is \(\lim_{h
ightarrow0}\frac{9e^{3h}}{10}\).

Step5: Evaluate the limit

Substitute \(h = 0\) into \(\frac{9e^{3h}}{10}\), we get \(\frac{9e^{0}}{10}=\frac{9\times1}{10}=\frac{9}{10}\).

Answer:

\(\frac{9}{10}\)