Let's solve each limit step-by-step.

1. $\lim_{{t \to +\infty}} \frac{1}{3} t^2$

   • As $t$ approaches infinity, $t^2$ also approaches infinity. Since there’s no factor that would make the expression finite or zero, this limit will be: $\lim_{{t \to +\infty}} \frac{1}{3} t^2 = +\infty$

2. $\lim_{{t \to +\infty}} (3e^t - 7)$

   • As $t$ approaches infinity, $e^t$ grows exponentially towards infinity. Thus, $3e^t$ will also tend to infinity, and subtracting 7 does not change this: $\lim_{{t \to +\infty}} (3e^t - 7) = +\infty$

3. $\lim_{{t \to -\infty}} \ln(7 - 3t)$

   • As $t$ approaches $-\infty$, $3t$ will approach $+\infty$ (since $-3$ times a large negative is positive). Consequently, $7 - 3t$ approaches $-\infty$, and the natural logarithm of a large negative value is undefined in real terms. Thus: $\lim_{{t \to -\infty}} \ln(7 - 3t) = \text{undefined}$

4. $\lim_{{t \to 0}} \ln(-3t)$

   • As $t$ approaches 0 from the positive side, $-3t$ approaches 0 from the negative side. However, the natural logarithm of a negative number is not defined in the real number system, so: $\lim_{{t \to 0}} \ln(-3t) = \text{undefined}$

5. $\lim_{{t \to 7/3}} \ln(3t - 7)$

   • As $t$ approaches $7/3$, $3t - 7$ approaches 0 from the positive side. The natural logarithm of a number approaching 0 from the positive side tends to $-\infty$: $\lim_{{t \to 7/3}} \ln(3t - 7) = -\infty$

6. $\lim_{{t \to 7/3}} \ln(7 - 3t)$

   • Similarly, as $t$ approaches $7/3$, $7 - 3t$ also approaches 0, but from the negative side. The natural logarithm of a negative value is undefined in the reals: $\lim_{{t \to 7/3}} \ln(7 - 3t) = \text{undefined}$

7. $\lim_{{t \to -\infty}} -\frac{1}{3} t^2$

   • Here, as $t$ approaches $-\infty$, $t^2$ approaches $+\infty$. Multiplying by $-\frac{1}{3}$ makes the expression approach $-\infty$: $\lim_{{t \to -\infty}} -\frac{1}{3} t^2 = -\infty$

8. $\lim_{{t \to +\infty}} -3t$

   • As $t$ approaches $+\infty$, $3t$ approaches $+\infty$, so $-3t$ will approach $-\infty$: $\lim_{{t \to +\infty}} -3t = -\infty$

9. $\lim_{{t \to +\infty}} -\frac{1}{3} t^2$

   • As $t$ approaches $+\infty$, $\frac{1}{3} t^2$ approaches $+\infty$, but with the negative sign in front, it will approach $-\infty$: $\lim_{{t \to +\infty}} -\frac{1}{3} t^2 = -\infty$

10. $\lim_{{t \to +\infty}} \ln(-3t)$

    • As $t$ approaches $+\infty$, $-3t$ approaches $-\infty$, making $\ln(-3t)$ undefined in the real number system: $\lim_{{t \to +\infty}} \ln(-3t) = \text{undefined}$

Would you like further explanations on any specific part of these solutions?

Related Questions

1. What are the rules for limits involving infinity?
2. How does the behavior of exponential functions affect limits?
3. Why is the natural logarithm undefined for negative values in real numbers?
4. How do we approach limits that result in undefined values?
5. What other techniques exist for solving complex limits?

Tip

When calculating limits, always check if the function approaches a finite value, infinity, or an undefined form as $t$ approaches the target value.