solve limit when t closes 1 on left side and the function is (arccos(t))/(t-1)

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.

Solve the limit of (arccos(t))/(t-1) as t approaches 1 from the left. This problem involves applying L'Hopital's Rule to an indeterminate form and differentiating inverse trigonometric functions. Suitable for undergraduate calculus students.

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