Math / Calculus

QuestionSubmit Answer 11. [-/1 Points] DETAILS MY NOTES SCALCET9M 4.4.015.
Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. $\lim _{t \rightarrow 0} \frac{e^{8 t}-1}{\sin (t)}$ $\square$ Need Help? Read It Watch It Submit Answer
12. [-/1 Points] DETAILS

Studdy Solution

STEP 1

Assumptions
1. We need to find the limit of the function as $t$ approaches 0.
2. The function is given by $\frac{e^{8t} - 1}{\sin(t)}$ .
3. We will use l'Hospital's Rule if the limit results in an indeterminate form like $\frac{0}{0}$ .
4. l'Hospital's Rule states that for limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , the limit of $\frac{f(t)}{g(t)}$ as $t \to a$ is equal to the limit of $\frac{f'(t)}{g'(t)}$ as $t \to a$ , provided this limit exists.

STEP 2

First, evaluate the limit directly by substituting $t = 0$ .
$\lim_{t \to 0} \frac{e^{8t} - 1}{\sin(t)} = \frac{e^{8 \cdot 0} - 1}{\sin(0)} = \frac{1 - 1}{0} = \frac{0}{0}$
Since this is an indeterminate form, we can apply l'Hospital's Rule.

STEP 3

Differentiate the numerator and the denominator separately.
The derivative of the numerator $e^{8t} - 1$ is $8e^{8t}$ .
The derivative of the denominator $\sin(t)$ is $\cos(t)$ .

STEP 4

Apply l'Hospital's Rule to find the limit.
$\lim_{t \to 0} \frac{e^{8t} - 1}{\sin(t)} = \lim_{t \to 0} \frac{8e^{8t}}{\cos(t)}$

STEP 5

Evaluate the limit by substituting $t = 0$ in the expression obtained after applying l'Hospital's Rule.
$\lim_{t \to 0} \frac{8e^{8t}}{\cos(t)} = \frac{8e^{8 \cdot 0}}{\cos(0)} = \frac{8 \cdot 1}{1} = 8$
The limit is 8.

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