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Find \(\lim _{x \rightarrow \infty} \sqrt{x^{2}+1}-\sqrt{x^{2}-1}\) \(\left(\right.\) Hint: Multiply by \(\left.\frac{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}+\sqrt{x^{2}-1}} .\right)\)

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Recognize the Indeterminate Form

Identify that directly substituting \(x \to \infty\) in \(\sqrt{x^2 + 1} - \sqrt{x^2 - 1}\) results in an indeterminate form of \(\infty - \infty\). To resolve this, we will use an algebraic trick called rationalizing.
02

Rationalize the Expression

Multiply the expression by a form of one to rationalize it: \(\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}\). This results in: \[\lim_{x \to \infty} \left(\sqrt{x^2 + 1} - \sqrt{x^2 - 1}\right) \cdot \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}.\]
03

Simplify Using Difference of Squares

Apply the difference of squares formula \((a-b)(a+b) = a^2 - b^2\):\[\lim_{x \to \infty} \frac{(x^2 + 1) - (x^2 - 1)}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}\].Simplify the numerator to get:\[\lim_{x \to \infty} \frac{2}{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}.\]
04

Evaluate the Limit as x Approaches Infinity

As \(x\) approaches infinity, the terms \(\sqrt{x^2 + 1}\) and \(\sqrt{x^2 - 1}\) dominate the equation. Approximating these terms, they are roughly \(x\). Thus the expression simplifies to:\[\lim_{x \to \infty} \frac{2}{x + x} = \lim_{x \to \infty} \frac{2}{2x} = \lim_{x \to \infty} \frac{1}{x} = 0\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
When dealing with limits in calculus, the term _indeterminate form_ may come up frequently. Indeterminate forms appear when the limit cannot be determined directly because substituting the value into the expression leads to uncertain results. In the context of the given problem, substituting \( x \to \infty \) directly results in an expression of the form \( \infty - \infty \). This is one of the classic indeterminate forms. The other common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \). Recognizing these forms is crucial, so we know alternative methods, like algebraic manipulations, are required to find the limit.
Rationalization Technique
Rationalization is a technique often used in calculus for simplifying the expressions involving roots. It involves multiplying the problematic expression by a cleverly chosen form of one, usually in the format \( \frac{A}{A} \), where \( A \) is the conjugate of the expression. In the given problem, we see this trick used to resolve the indeterminate form.

In the expression \( \sqrt{x^2+1} - \sqrt{x^2-1} \), multiplying by \( \frac{\sqrt{x^2+1} + \sqrt{x^2-1}}{\sqrt{x^2+1} + \sqrt{x^2-1}} \) rationalizes it. This effectively transforms the expression into a difference of squares format, simplifying the problem. By eliminating the square roots, the expression becomes easier to handle, enabling us to evaluate the limit more feasibly.
Difference of Squares
The difference of squares is a straightforward yet powerful algebraic method expressed as \( a^2 - b^2 = (a-b)(a+b) \). This identity is instrumental in rationalizing expressions that involve roots. When applied to the rationalized form of the problem \( (\sqrt{x^2+1} - \sqrt{x^2-1})(\sqrt{x^2+1} + \sqrt{x^2-1}) \), it simplifies the numerator to \((x^2+1) - (x^2-1) = 2\).

The difference of squares effectively eliminates the square roots, which might otherwise complicate the limit calculation. This simplification allows us to focus on the behavior of the denominator as \( x \) approaches infinity.
Infinity in Calculus
Infinity often represents a challenging concept in calculus. It is not a number, but a notion that describes the unbounded nature of values. Calculating limits involving infinity, such as \( \lim_{x \to \infty} \), requires a clear understanding of how functions behave as they grow unceasingly.

In our problem, the core challenge was understanding that as \( x \) grows very large, \( \sqrt{x^2 + 1} \) and \( \sqrt{x^2 - 1} \) both approach \( x \). This allows us to approximate the expression in the limit:
\( \frac{2}{\sqrt{x^2+1} + \sqrt{x^2-1}} \) further simplifies to \( \frac{2}{2x} \), which then approaches zero because \( \frac{1}{x} \to 0 \) as \( x \to \infty \).

This insight into the nature of infinity and how it interacts with functions is crucial in mastering limits and calculus as a whole.

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