Problem 57 - Calculus Early Transcendentals

Question

$$ \lim _{x \rightarrow \infty} f(x) \text { if, for all } x>1 $$ $$ \frac{10 \mathrm{e}^{x}-21}{2 \mathrm{e}^{\mathrm{x}}}<\mathrm{f}(\mathrm{x})<\frac{5 \sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}-1}} $$

Step-by-Step Solution

Answer

The limit is 5 as \( x \to \infty \).

Step 1: Understand the Exercise

We are given that for all \( x > 1 \), the function \( f(x) \) is bounded between two expressions, \( \frac{10 e^x - 21}{2 e^x} \) and \( \frac{5 \sqrt{x}}{\sqrt{x-1}} \). We need to find \( \lim_{x \to \infty} f(x) \).

Step 2: Solve the Lower Bound

We first consider the lower bound of \( f(x) \), which is \( \frac{10 e^x - 21}{2 e^x} \). Simplifying, we have:\[\frac{10 e^x - 21}{2 e^x} = \frac{10 e^x}{2e^x} - \frac{21}{2e^x} = 5 - \frac{21}{2e^x}.\]As \( x \to \infty \), \( \frac{21}{2e^x} \to 0 \). Therefore:\[\lim_{x \to \infty} \frac{10 e^x - 21}{2 e^x} = 5.\]

Step 3: Solve the Upper Bound

Next, consider the upper bound \( \frac{5\sqrt{x}}{\sqrt{x-1}} \). Factoring within the square roots, we have:\[\frac{5\sqrt{x}}{\sqrt{x-1}} = 5 \sqrt{\frac{x}{x-1}}.\]As \( x \to \infty \), the expression \( \frac{x}{x-1} \) approaches \( 1 \), so \( \sqrt{\frac{x}{x-1}} \to 1 \). Therefore:\[\lim_{x \to \infty} \frac{5\sqrt{x}}{\sqrt{x-1}} = 5.\]

Step 4: Evaluate the Limit of f(x)

Since \( f(x) \) is always between \( \frac{10 e^x - 21}{2 e^x} \) and \( \frac{5\sqrt{x}}{\sqrt{x-1}} \), both of which approach 5 as \( x \to \infty \), by the Squeeze Theorem, we conclude:\[\lim_{x \to \infty} f(x) = 5.\]

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus for finding the limit of a function. When you have a function trapped between two other functions, and if both the upper and lower bounding functions approach the same limit when x goes to infinity, you can conclude that the function in between must also approach this limit.
In our exercise, we have a function \( f(x) \) that is squeezed between two functions: \( \frac{10 e^x - 21}{2 e^x} \) and \( \frac{5 \sqrt{x}}{\sqrt{x-1}} \). As both approaches the limit 5 as \( x \) becomes infinitely large, it allows us to conclude that \( \lim_{x \to \infty} f(x) = 5 \).
This theorem simplifies problems where evaluating the limit of a function directly is complex, providing a shortcut to finding the limit by using simpler boundary functions.

Exponential Functions

Exponential functions, like \( e^x \), are incredibly important in calculus and real-world applications. These functions grow very rapidly as \( x \) increases, which is why they show up frequently in problems related to limits at infinity.
An exponential function has the form \( a^x \), where \( e \) (approximately 2.718) is a commonly used base due to its wonderful properties in calculus. It grows faster than any polynomial, which is why terms involving exponentials dominate in limits involving large \( x \).
In the given problem, the presence of \( e^x \) in the numerator and denominator helps us factor and simplify the expressions in finding limits.

Limits at Infinity

Limits at infinity are used to understand the behavior of functions as they let their variable, like \( x \), go off to infinity or negative infinity. This is incredibly useful in many fields of mathematics, helping us predict long-term behavior.
In this exercise, we are asked to find \( \lim_{x \to \infty} f(x) \). As \( x \) becomes very large, the values of \( f(x) \), or its boundary functions, \( \frac{10 e^x - 21}{2 e^x} \) and \( \frac{5 \sqrt{x}}{\sqrt{x-1}} \), stabilize to a common value, which is 5.
This illustrates how we can use calculus tools to analyze functions’ end behavior and direction as variables move towards infinity.

Square Roots in Limits

Square roots can transform expressions and play a significant role in limits, especially when nested within functions. These transformations can actually simplify an otherwise complicated limit calculation.
In this problem, \( \frac{5 \sqrt{x}}{\sqrt{x-1}} \) involves a square root in the denominator. By factoring the expression inside the square root and simplifying, we see that as \( x \to \infty \), \( \sqrt{\frac{x}{x-1}} \to 1 \), indicating that the effect of the square roots diminishes.
Hence, in limits, square roots often simplify as you evaluate them at infinity, which can be a turning point to simplifying a complex limit evaluation.