Problem 33

Question

The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of $x :$ Divide numerator and denominator by the highest power of $x$ in the denominator and proceed from there. Find the limits in Exercises $23-36$. $$\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}+1}}{x+1}$$

Step-by-Step Solution

Verified

Answer

The limit is 1.

1Step 1: Identify the Highest Power of x

In the expression $ \frac{\sqrt{x^{2}+1}}{x+1} $, the highest power of $ x $ in the denominator is $ x^1 $. We will divide both the numerator and the denominator by $ x^1 $.

2Step 2: Simplify the Expression

Divide both the numerator and the denominator by $ x $. The expression becomes: \[ \frac{\frac{\sqrt{x^2 + 1}}{x}}{\frac{x + 1}{x}} = \frac{\sqrt{1 + \frac{1}{x^2}}}{1 + \frac{1}{x}}. \]

3Step 3: Evaluate the Limit as x Approaches Infinity

As $ x \rightarrow \infty $, both $ \frac{1}{x} $ and $ \frac{1}{x^2} $ approach 0. Thus, the expression simplifies to: \[ \frac{\sqrt{1 + 0}}{1 + 0} = \frac{1}{1} = 1. \]

Key Concepts

Dividing by Highest Power of xEvaluating Limits as x Approaches InfinitySimplifying Expressions with Square Roots

Dividing by Highest Power of x

When evaluating the limit of a rational function like the one we have in the given problem, we start by dividing both the numerator and the denominator by the highest power of $x$ found in the denominator. In this example, the highest power of $x$ in the denominator is $x^1$. By dividing each term by $x^1$, we simplify the expression, making it easier to determine its behavior as $x$ approaches infinity. This step is essential because it helps to eliminate the dominant factor of $x$ in each part of the expression.

Numerator: Divide $\sqrt{x^2 + 1}$ by $x$ resulting in $\sqrt{1 + \frac{1}{x^2}}$.
Denominator: Divide $x + 1$ by $x$, resulting in $1 + \frac{1}{x}$.

After these transformations, the complex original fraction is transformed into a much simpler form: $\frac{\sqrt{1 + \frac{1}{x^2}}}{1 + \frac{1}{x}}$. This makes the next steps of the process far more manageable as we prepare to evaluate the limit.

Evaluating Limits as x Approaches Infinity

Once the fractions are simplified by dividing by the highest power of $x$, we can focus on evaluating the behavior of the expression as $x$ approaches infinity. Primarily, expressions like $\frac{1}{x}$ and $\frac{1}{x^2}$ are vital because, as $x$ becomes very large, both fractions tend toward zero:

$\frac{1}{x}$: As $x$ gets larger, the value of $\frac{1}{x}$ diminishes, practically approaching zero.
$\frac{1}{x^2}$: Similarly, $\frac{1}{x^2}$ heads towards zero even faster, due to the squared term.

This means that the simplified expression $\frac{\sqrt{1 + \frac{1}{x^2}}}{1 + \frac{1}{x}}$ will simplify further:- The numerator $\sqrt{1 + \frac{1}{x^2}}$ simplifies to $\sqrt{1 + 0}$, which equals 1.- The denominator $1 + \frac{1}{x}$ simplifies to $1 + 0$, which also equals 1.Thus, as $x$ approaches infinity, the entire expression converges to $\frac{1}{1} = 1$. This final step gives us the limit of the original function.

Simplifying Expressions with Square Roots

When working with square roots within limits, it's important to manipulate the expression in a way that clarifies the behavior of the square root as $x$ changes. In our sample problem, we start by examining $\sqrt{x^2 + 1}$ in the numerator. By dividing $\sqrt{x^2 + 1}$ by $x$, it becomes $\sqrt{1 + \frac{1}{x^2}}$. This transformation allows us to isolate terms that approach zero as $x$ goes to infinity, simplifying the limit process.The square root of any expression adds a layer of complexity because of the non-linearity of square root operations. However, when expressions are structured like $\sqrt{1 + \frac{1}{x^2}}$, the limit operation becomes straightforward:

Focus on the dominant constant or term within the root (here, it's 1). The $\frac{1}{x^2}$ becomes negligible as $x$ grows.
Consequently, $\sqrt{1 + 0} = 1$.

The process of reducing complex expressions under square roots to simpler forms as variables approach infinity can greatly simplify any calculations involved in limit problems.

Problem 33

Other exercises in this chapter

Problem 32

Each of Exercises $31-36$ gives a function $f(x),$ a point $c,$ and a positive number $\epsilon .$ Find $L=\lim _{x \rightarrow c} f(x) .$ Then find a

View solution

Problem 33

Find the limits. Are the functions continuous at the point being approached? $$\lim _{y \rightarrow 1} \sec \left(y \sec ^{2} y-\tan ^{2} y-1\right)$$

View solution

Problem 33

Find the limits in Exercises $21-42$ $$\lim _{x \rightarrow 0} \frac{x-x \cos x}{\sin ^{2} 3 x}$$

View solution

Problem 33

Limits of quotients Find the limits in Exercises $23-42$ $$\lim _{u \rightarrow 1} \frac{u^{4}-1}{u^{3}-1}$$

View solution