Calculus worksheet on limits at infinity with handwritten solutions and explanations.

Problem Analysis and Solution

The image contains problems related to limits at infinity. Let's solve each problem step by step, explaining the reasoning behind each solution.

---

#### Problem 5: Find each limit if possible

We are given three limits to evaluate:

1. \( \lim_{x \to \infty} \frac{x^9 + x}{x^9 - 6} \)
2. \( \lim_{x \to \infty} \frac{x^9 + x}{x^{10} - 6} \)
3. \( \lim_{x \to -\infty} \frac{x^9 + x}{x^9 - 6} \)

##### Part (a): \( \lim_{x \to \infty} \frac{x^9 + x}{x^9 - 6} \)

To evaluate this limit, we divide the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^9 \):

\[
\lim_{x \to \infty} \frac{x^9 + x}{x^9 - 6} = \lim_{x \to \infty} \frac{\frac{x^9}{x^9} + \frac{x}{x^9}}{\frac{x^9}{x^9} - \frac{6}{x^9}}
\]

Simplify each term:

\[
= \lim_{x \to \infty} \frac{1 + \frac{1}{x^8}}{1 - \frac{6}{x^9}}
\]

As \( x \to \infty \), the terms \( \frac{1}{x^8} \) and \( \frac{6}{x^9} \) approach 0. Therefore:

\[
= \frac{1 + 0}{1 - 0} = 1
\]

So, the limit is:

\[
\boxed{1}
\]

##### Part (b): \( \lim_{x \to \infty} \frac{x^9 + x}{x^{10} - 6} \)

Again, we divide the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x^{10} \):

\[
\lim_{x \to \infty} \frac{x^9 + x}{x^{10} - 6} = \lim_{x \to \infty} \frac{\frac{x^9}{x^{10}} + \frac{x}{x^{10}}}{\frac{x^{10}}{x^{10}} - \frac{6}{x^{10}}}
\]

Simplify each term:

\[
= \lim_{x \to \infty} \frac{\frac{1}{x} + \frac{1}{x^9}}{1 - \frac{6}{x^{10}}}
\]

As \( x \to \infty \), the terms \( \frac{1}{x} \), \( \frac{1}{x^9} \), and \( \frac{6}{x^{10}} \) approach 0. Therefore:

\[
= \frac{0 + 0}{1 - 0} = 0
\]

So, the limit is:

\[
\boxed{0}
\]

##### Part (c): \( \lim_{x \to -\infty} \frac{x^9 + x}{x^9 - 6} \)

For this limit, we again divide the numerator and the denominator by \( x^9 \):

\[
\lim_{x \to -\infty} \frac{x^9 + x}{x^9 - 6} = \lim_{x \to -\infty} \frac{\frac{x^9}{x^9} + \frac{x}{x^9}}{\frac{x^9}{x^9} - \frac{6}{x^9}}
\]

Simplify each term:

\[
= \lim_{x \to -\infty} \frac{1 + \frac{1}{x^8}}{1 - \frac{6}{x^9}}
\]

As \( x \to -\infty \), the terms \( \frac{1}{x^8} \) and \( \frac{6}{x^9} \) approach 0. However, note that \( x^9 \) is negative when \( x \) is negative, so the sign of the terms involving \( x^9 \) must be considered. But since \( \frac{1}{x^8} \) and \( \frac{6}{x^9} \) still approach 0, the limit is:

\[
= \frac{1 + 0}{1 - 0} = 1
\]

So, the limit is:

\[
\boxed{1}
\]

---

#### Problem 6: Find the limit, if it exists

We are given several limits to evaluate. Let's solve them one by one.

##### Part (a): \( \lim_{t \to \infty} \sqrt{t^2 + t} - t \)

To evaluate this limit, we rationalize the expression by multiplying and dividing by the conjugate:

\[
\lim_{t \to \infty} \left( \sqrt{t^2 + t} - t \right) = \lim_{t \to \infty} \frac{\left( \sqrt{t^2 + t} - t \right) \left( \sqrt{t^2 + t} + t \right)}{\sqrt{t^2 + t} + t}
\]

The numerator simplifies as follows:

\[
\left( \sqrt{t^2 + t} \right)^2 - t^2 = (t^2 + t) - t^2 = t
\]

So the expression becomes:

\[
\lim_{t \to \infty} \frac{t}{\sqrt{t^2 + t} + t}
\]

Divide the numerator and the denominator by \( t \):

\[
= \lim_{t \to \infty} \frac{\frac{t}{t}}{\frac{\sqrt{t^2 + t}}{t} + \frac{t}{t}} = \lim_{t \to \infty} \frac{1}{\sqrt{1 + \frac{1}{t}} + 1}
\]

As \( t \to \infty \), \( \frac{1}{t} \to 0 \). Therefore:

\[
= \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}
\]

So, the limit is:

\[
\boxed{\frac{1}{2}}
\]

##### Part (b): \( \lim_{t \to \infty} \sqrt{t^2 + t} + t \)

This limit diverges to infinity because both \( \sqrt{t^2 + t} \) and \( t \) grow without bound as \( t \to \infty \). Therefore:

\[
\boxed{\infty}
\]

##### Part (c): \( \lim_{t \to -\infty} \sqrt{t^2 + t} - t \)

For \( t \to -\infty \), we need to be careful with the sign of \( t \). Rewrite \( t \) as \( -|t| \):

\[
\lim_{t \to -\infty} \left( \sqrt{t^2 + t} - t \right) = \lim_{t \to -\infty} \left( \sqrt{t^2 + t} + |t| \right)
\]

Since \( t \to -\infty \), \( |t| = -t \). Rationalize the expression:

\[
\lim_{t \to -\infty} \left( \sqrt{t^2 + t} + (-t) \right) = \lim_{t \to -\infty} \frac{\left( \sqrt{t^2 + t} + (-t) \right) \left( \sqrt{t^2 + t} - (-t) \right)}{\sqrt{t^2 + t} - (-t)}
\]

The numerator simplifies as follows:

\[
\left( \sqrt{t^2 + t} \right)^2 - (-t)^2 = (t^2 + t) - t^2 = t
\]

So the expression becomes:

\[
\lim_{t \to -\infty} \frac{t}{\sqrt{t^2 + t} - (-t)} = \lim_{t \to -\infty} \frac{t}{\sqrt{t^2 + t} + t}
\]

Divide the numerator and the denominator by \( |t| = -t \):

\[
= \lim_{t \to -\infty} \frac{\frac{t}{-t}}{\frac{\sqrt{t^2 + t}}{-t} + \frac{t}{-t}} = \lim_{t \to -\infty} \frac{-1}{\sqrt{1 + \frac{1}{t}} - 1}
\]

As \( t \to -\infty \), \( \frac{1}{t} \to 0 \). Therefore:

\[
= \frac{-1}{\sqrt{1 + 0} - 1} = \frac{-1}{1 - 1} = \frac{-1}{0^+} = -\infty
\]

So, the limit is:

\[
\boxed{-\infty}
\]

##### Part (d): \( \lim_{t \to -\infty} \sqrt{t^2 + t} + t \)

For \( t \to -\infty \), we have:

\[
\lim_{t \to -\infty} \left( \sqrt{t^2 + t} + t \right)
\]

Rationalize the expression:

\[
\lim_{t \to -\infty} \left( \sqrt{t^2 + t} + t \right) = \lim_{t \to -\infty} \frac{\left( \sqrt{t^2 + t} + t \right) \left( \sqrt{t^2 + t} - t \right)}{\sqrt{t^2 + t} - t}
\]

The numerator simplifies as follows:

\[
\left( \sqrt{t^2 + t} \right)^2 - t^2 = (t^2 + t) - t^2 = t
\]

So the expression becomes:

\[
\lim_{t \to -\infty} \frac{t}{\sqrt{t^2 + t} - t}
\]

Divide the numerator and the denominator by \( |t| = -t \):

\[
= \lim_{t \to -\infty} \frac{\frac{t}{-t}}{\frac{\sqrt{t^2 + t}}{-t} - \frac{t}{-t}} = \lim_{t \to -\infty} \frac{-1}{\sqrt{1 + \frac{1}{t}} - 1}
\]

As \( t \to -\infty \), \( \frac{1}{t} \to 0 \). Therefore:

\[
= \frac{-1}{\sqrt{1 + 0} - 1} = \frac{-1}{1 - 1} = \frac{-1}{0^-} = \infty
\]

So, the limit is:

\[
\boxed{\infty}
\]

---

Final Answers

1. \( \lim_{x \to \infty} \frac{x^9 + x}{x^9 - 6} = \boxed{1} \)
2. \( \lim_{x \to \infty} \frac{x^9 + x}{x^{10} - 6} = \boxed{0} \)
3. \( \lim_{x \to -\infty} \frac{x^9 + x}{x^9 - 6} = \boxed{1} \)
4. \( \lim_{t \to \infty} \sqrt{t^2 + t} - t = \boxed{\frac{1}{2}} \)
5. \( \lim_{t \to \infty} \sqrt{t^2 + t} + t = \boxed{\infty} \)
6. \( \lim_{t \to -\infty} \sqrt{t^2 + t} - t = \boxed{-\infty} \)
7. \( \lim_{t \to -\infty} \sqrt{t^2 + t} + t = \boxed{\infty} \)