We are tasked with finding the following limit:

$\lim_{t \to 1^-} \frac{\arccos(t)}{t - 1}$

Step 1: Behavior of $\arccos(t)$ as $t \to 1^{-}$

The function $\arccos(t)$ represents the inverse cosine, which approaches 0 as $t$ approaches 1 from the left. Specifically:

$\arccos(1) = 0$

Thus, as $t \to 1^-$, $\arccos(t) \to 0$. This makes the expression of the form $\frac{0}{0}$, which is indeterminate. We can now apply L'Hopital's Rule, since we have a $\frac{0}{0}$ indeterminate form.

Step 2: Applying L'Hopital's Rule

L'Hopital's Rule states that for an indeterminate form $\frac{0}{0}$, we can evaluate the limit by differentiating the numerator and denominator:

[ \lim_{t \to 1^-} \frac{\arccos(t)}{t - 1} = \lim_{t \to 1^-} \frac{\frac{d}{dt}[\arccos(t)]}{\frac{d}{dt}[t - 1]} ]

Derivative of the numerator:

The derivative of $\arccos(t)$ is:

$\frac{d}{dt}[\arccos(t)] = \frac{-1}{\sqrt{1 - t^2}}$

Derivative of the denominator:

The derivative of $t - 1$ is simply:

$\frac{d}{dt}[t - 1] = 1$

Step 3: Simplifying the expression

Now, substituting these derivatives into the limit:

$\lim_{t \to 1^-} \frac{-1/\sqrt{1 - t^2}}{1} = \lim_{t \to 1^-} \frac{-1}{\sqrt{1 - t^2}}$

As $t \to 1^-$, $1 - t^2$ approaches $0^+$. Therefore, the square root $\sqrt{1 - t^2}$ also approaches 0 from the positive side. Hence, the limit becomes:

$\lim_{t \to 1^-} \frac{-1}{\sqrt{1 - t^2}} = -\infty$

Conclusion:

The limit is:

$\lim_{t \to 1^-} \frac{\arccos(t)}{t - 1} = -\infty$

Would you like further details or clarification on any step?

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Here are 5 related questions you may want to explore:

1. What is L'Hopital's Rule and how does it apply to indeterminate forms?
2. How do we differentiate inverse trigonometric functions, like $\arccos(t)$?
3. What are the conditions under which L'Hopital's Rule can be used?
4. How does the behavior of $\arccos(t)$ change as $t$ approaches $-1$?
5. What happens to the limit if $t \to 1^+$ instead of $t \to 1^-$?

Tip: When dealing with limits resulting in indeterminate forms like $\frac{0}{0}$, always consider applying L'Hopital's Rule.