Problem 38 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}-1}\right) $$

Step-by-Step Solution

Answer

The limit is 0 as \(x\) approaches infinity.

Step 1: Identify the Indeterminate Form

First, identify that the expression \(x - \sqrt{x^2 - 1}\) becomes indeterminate of the form \(\infty - \infty\) as \(x\) approaches infinity.

Step 2: Rationalization

To resolve the indeterminate form, multiply the expression by the conjugate \( \frac{(x + \sqrt{x^2 - 1})}{(x + \sqrt{x^2 - 1})} \): \[\lim _{x \rightarrow \infty} \left(\frac{(x - \sqrt{x^2 - 1})(x + \sqrt{x^2 - 1})}{x + \sqrt{x^2 - 1}}\right)\]

Step 3: Simplify the Expression

After multiplying, the numerator simplifies to:\[(x^2 - (x^2 - 1)) = 1\]Thus, the limit becomes:\[\lim _{x \rightarrow \infty} \frac{1}{x + \sqrt{x^2 - 1}}\]

Step 4: Evaluate the Limit

As \(x\) approaches infinity, the denominator \(x + \sqrt{x^2 - 1}\) approaches \(2x\). Thus, the limit becomes:\[\lim _{x \rightarrow \infty} \frac{1}{2x} = 0\]

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms occur when evaluating limits, making it challenging to immediately determine their values. An expression becomes indeterminate when direct substitution into a limit gives us an undefined or uncertain form. These forms include \(0/0\), \(\infty/\infty\), \(\infty - \infty\), among others. In the given exercise, the expression \(x - \sqrt{x^2 - 1}\) results in \(\infty - \infty\) as \(x\) approaches infinity. This is a classic case of an indeterminate form. To manage indeterminate forms, mathematicians often employ techniques like l'Hospital's Rule or algebraic manipulation, ensuring they can find meaningful limit values in calculus problems.

Rationalization

Rationalization is a technique used to simplify expressions, particularly useful for dealing with indeterminate forms. It involves multiplying an expression by a conjugate to remove square roots or complex fractions. For the problem at hand, when determining \(\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}-1})\), rationalization helps eliminate the indeterminate \(\infty - \infty\) form. By multiplying with the conjugate \(\frac{(x + \sqrt{x^2 - 1})}{(x + \sqrt{x^2 - 1})}\), the expression simplifies significantly. This technique reduces the initial problem to a much easier form without square roots, helping us proceed with calculating limits more straightforwardly. Rationalization is a powerful tool, particularly when dealing with radical expressions.

Limits in Calculus

Limits represent a fundamental concept in calculus, describing the behavior of functions as they approach specific points or infinity. In the exercise provided, we're concerned with the limit as \(x\) approaches infinity. By simplifying using rationalization, you've reduced the expression to \(\frac{1}{x + \sqrt{x^2 - 1}}\). As \(x\) becomes infinitely large, the dominant term in the denominator becomes \(x\) itself, which simplifies the limit calculation. The denominator \(x + \sqrt{x^2 - 1}\) approaches \(2x\), allowing the entire expression \(\frac{1}{2x}\) to approach zero. Limits like these showcase why understanding the behavior at infinity and recognizing dominant terms is crucial, helping verify that our calculated limits are reasonable and correct.