Problem 84 - Calculus Single Variable

Question

Evaluate $$ \lim _{x \rightarrow \infty} \frac{x^{(x+1) / x}}{\sqrt{1+x^{2}}} $$ .

Step-by-Step Solution

Answer

The limit evaluates to 1.

Step 1: Rewrite the Expression

First, we need to rewrite the given expression in a more workable form. Let's rewrite the expression inside the limit as \( \frac{x^{\frac{x+1}{x}}}{\sqrt{1+x^2}} \). This simplifies to \( \frac{x^{1 + \frac{1}{x}}}{\sqrt{1+x^2}} \).

Step 2: Analyze the Numerator

The term \( x^{1 + \frac{1}{x}} \) can be split using the properties of exponentials: \( x^{1 + \frac{1}{x}} = x^1 \cdot x^{\frac{1}{x}} \). Now, \( x^{\frac{1}{x}} = e^{\ln(x)^{1/x}} \approx e^0 = 1 \) as \( x \to \infty \), so the numerator becomes approximately \( x \).

Step 3: Simplify the Denominator

Now consider the denominator \( \sqrt{1+x^2} \). For large \( x \), the term \( x^2 \) dominates over \( 1 \). Thus, \( \sqrt{1+x^2} \approx \sqrt{x^2} = x \).

Step 4: Form the Limit Expression

Putting it all together, the expression inside the limit simplifies to \( \frac{x \cdot 1}{x} = 1 \) as \( x \to \infty \). This means the limit can be computed as just \( \lim_{x \to \infty} 1 = 1 \).

Step 5: Evaluate the Limit

Since we've simplified the expression inside the limit to \( 1 \), the limit evaluates to \( 1 \). Thus, we have \( \lim _{x \rightarrow \infty} \frac{x^{(x+1) / x}}{\sqrt{1+x^{2}}} = 1 \).

Key Concepts

Exponentials

Exponentials are mathematical functions where a base number is raised to a power or exponent. In our problem, the exponential is expressed as \( x^{1 + \frac{1}{x}} \). This illustrates the ideas of both base and exponent:

• The base here is \( x \).
• The exponent is \( 1 + \frac{1}{x} \).

To simplify this, we used the property that allows us to split exponents. This means we can write it as \( x^1 \cdot x^{\frac{1}{x}} \). As \( x \) approaches infinity, \( x^{\frac{1}{x}} \) approaches \( e^{\ln(x)^{1/x}} \), which simplifies to approximately 1. This is because the natural logarithm \( \ln(x) \) grows slower than a linear rate, causing the exponent \( \frac{1}{x} \) to make the overall expression tend toward 0, resulting in \( e^0 \).Understanding exponentials is key to rewriting the original expression so the limit can be evaluated. They allow us to manipulate and simplify the expression more effectively.

Square Roots

Square roots involve finding a number that, when squared, gives the original number. In the context of our problem, we're dealing with the square root in the denominator: \( \sqrt{1 + x^2} \).

• Here, for huge values of \( x \), the term \( x^2 \) becomes significantly larger than 1.
• This allows us to make an approximation: \( \sqrt{1+x^2} \approx \sqrt{x^2} \).
• Thus, simplifying to \( x \).

The simplifying assumption that \( x^2 \) overwhelms the 1 term is central to determining the limit. Thus, the dominantly large \( x^2 \) ensures \( \sqrt{1+x^2} \) is approximately equal to \( x \) as \( x \) becomes extremely large. Knowing these principles helps to focus on the essential parts of the expression and disregard negligible terms under infinite limits.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as it approaches a particular value, often infinity. With limits, understanding asymptotic behavior is crucial.

• The asymptotic behavior provided us insight when \( x \rightarrow \infty \) in analyzing both the numerator and the denominator.
• The term \( x^{\frac{1}{x}} \approx 1 \) becomes crucial. As \( x \) goes to infinity, the exponent goes to zero.
• Similarly, the denominator \( \sqrt{1+x^2} \approx x \) shows asymptotic behavior as terms that grow slower can be ignored.

Grasping asymptotic behavior helps refine which function components have meaningful impact as the variable grows. This understanding allows simplifying the limit expression for objects like the numerator and denominator to make them simpler or even determinate under limit evaluation.

Properties of Logarithms

Logarithms help us convert multiplication into addition which is often easier to manage in limits. Here, properties of logarithms play a subtle but significant role in understanding the expression \( x^{1 + \frac{1}{x}} \).

• Using the property \( x^a = e^{a \ln(x)} \), we manipulate the expression \( x^{\frac{1}{x}} = e^{\ln(x)^{1/x}} \).
• As \( x \to \infty \), the natural log \( \ln(x) \) grows slower than the terms \( 1/x \), smoothing out the growth rate tending the exponent towards zero.
• This results in \( e^0 = 1 \), simplifying the entire exponent part. The real power here is that properties of logarithms help to untangle complex exponential behavior under limit operations.

Logarithmic transformations are vital when dealing with exponentials approaching a limit, as they provide a clearer path to simplification and understanding limit behavior.