To determine if the given functions are continuous at $t_0 = 0$, we need to check whether the function approaches a finite limit as $t$ approaches 0 and if the function itself is defined at $t_0 = 0$.

1. $F_1(t) = \frac{e^t - 1}{t}$

This function has a potential issue at $t = 0$ because the denominator becomes 0. So, we need to evaluate the limit as $t \to 0$.

$\lim_{t \to 0} \frac{e^t - 1}{t}$

This is a standard limit, and using L'Hopital's Rule (since it's of the form $\frac{0}{0}$):

$\lim_{t \to 0} \frac{e^t - 1}{t} = \lim_{t \to 0} \frac{e^t}{1} = e^0 = 1$

Thus, the limit exists and equals 1. We can now define $F_1(0) = 1$ to make the function continuous at $t_0 = 0$.

2. $F_2(t) = \tan(t)$

The function $\tan(t)$ is defined and continuous everywhere except at points where $t = \frac{\pi}{2} + n\pi$, where $n$ is an integer. Since 0 is not one of these points, the function $\tan(t)$ is continuous at $t_0 = 0$, and we have:

$\lim_{t \to 0} \tan(t) = \tan(0) = 0$

Thus, $\tan(t)$ is continuous at $t_0 = 0$.

3. $F_3(t) = \frac{1}{t + 1}$

This function is defined and continuous for all real values of $t$. At $t = 0$:

$F_3(0) = \frac{1}{0 + 1} = 1$

Thus, $F_3(t)$ is continuous at $t_0 = 0$.

Conclusion:

• $F_1(t) = \frac{e^t - 1}{t}$ is continuous at $t_0 = 0$ if we define $F_1(0) = 1$.
• $F_2(t) = \tan(t)$ is continuous at $t_0 = 0$.
• $F_3(t) = \frac{1}{t + 1}$ is continuous at $t_0 = 0$.

Would you like any further details or explanations?

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Here are 5 related questions:

1. How do you apply L'Hopital's Rule to other limits?
2. Can you find the derivative of $F_1(t) = \frac{e^t - 1}{t}$?
3. What is the behavior of $\tan(t)$ at its discontinuities?
4. How do you verify continuity for more complicated functions at a point?
5. What are the conditions for a function to be continuous at a point?

Tip: To check continuity, always verify that the function is defined at the point and that the limit as $t \to t_0$ equals the function's value at $t_0$.