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It appears that you've uploaded an image of a limit problem. The expression in the image seems to be:

\[
\lim_{{x \to \infty}} \frac{x^2 + 1}{\sqrt{x^4 - 2}}
\]

Let me solve this step by step:

### Step 1: Simplify the expression
We first divide both the numerator and denominator by \(x^2\), the highest power of \(x\) in the denominator inside the square root.

\[
\lim_{{x \to \infty}} \frac{x^2 + 1}{\sqrt{x^4 - 2}} = \lim_{{x \to \infty}} \frac{x^2(1 + \frac{1}{x^2})}{x^2\sqrt{1 - \frac{2}{x^4}}}
\]

### Step 2: Cancel out common terms
The \(x^2\) terms cancel out in both the numerator and denominator:

\[
\lim_{{x \to \infty}} \frac{1 + \frac{1}{x^2}}{\sqrt{1 - \frac{2}{x^4}}}
\]

### Step 3: Evaluate the limit
As \(x \to \infty\), \(\frac{1}{x^2} \to 0\) and \(\frac{2}{x^4} \to 0\). This simplifies the expression to:

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

### Final Answer:
\[
\lim_{{x \to \infty}} \frac{x^2 + 1}{\sqrt{x^4 - 2}} = 1
\]

Would you like more details or have any other questions? Here are some related questions for further practice:

1. What is the limit of \(\lim_{{x \to \infty}} \frac{x^2 - 1}{\sqrt{x^4 + 3x^2}}\)?
2. How does L'Hopital's rule apply to limits of indeterminate forms?
3. How would you handle the limit if the expression involved a negative infinity?
4. What is the importance of dividing by the highest power of \(x\) in limits at infinity?
5. What happens if we change the limit to \(x \to 0\) instead of \(x \to \infty\)?

**Tip**: When simplifying limits at infinity, always factor out the highest power of \(x\) to handle expressions more easily!

Find the limit of (x^2 + 1) / sqrt(x^4 - 2) as x approaches infinity.

Learn how to evaluate the limit of (x^2 + 1) divided by sqrt(x^4 - 2) as x tends to infinity. This solution involves simplifying expressions with dominant terms and applying properties of limits. Ideal for high school and early college-level students.

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