https://file.aiquickdraw.com/mathbot/file/1726518180260xx9z1neg.jpg

The expression given in the image is:

\[
\lim_{x \to \infty} \left( 2x^{2/3} - 7x^{-1} \right)
\]

Let's evaluate this limit step by step.

### Step 1: Analyze the behavior of each term as \( x \to \infty \)
- The term \( 2x^{2/3} \):
  - As \( x \to \infty \), since \( \frac{2}{3} > 0 \), \( x^{2/3} \to \infty \), and multiplying by 2 does not change the behavior. So, \( 2x^{2/3} \to \infty \).

- The term \( -7x^{-1} \):
  - As \( x \to \infty \), since \( -1 < 0 \), \( x^{-1} \to 0 \). Multiplying by -7, we get \( -7x^{-1} \to 0 \).

### Step 2: Combine the terms
The limit becomes:
\[
\lim_{x \to \infty} \left( 2x^{2/3} - 7x^{-1} \right) = \infty - 0 = \infty
\]

Thus, the limit is:

\[
\boxed{\infty}
\]

Would you like any further explanation or have any other questions?

Here are five related questions you might find helpful:
1. How does the behavior of exponents affect limits as \( x \to \infty \)?
2. What are some common techniques for solving limits involving powers of \( x \)?
3. How would this limit change if the exponent of \( x \) were negative in both terms?
4. Can you explain how L'Hopital's Rule might apply to limits like this one?
5. What would the limit be if the exponents were larger than 1?

**Tip:** When handling limits at infinity, pay close attention to the exponents in each term. Terms with higher positive exponents dominate the behavior.

lim_{x \to \infty} \left( 2x^{2/3} - 7x^{-1} \right)

Learn how to solve the limit problem involving exponents, lim (2x^(2/3) - 7x^(-1)) as x approaches infinity. This detailed step-by-step solution focuses on analyzing the behavior of terms with fractional and negative exponents. Suitable for students in Grades 11-12.

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