Problem 51 - Single Variable Calculus

Question

Evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}+1}\right) $$

Step-by-Step Solution

Answer

Using l'Hôpital's Rule, the limit \(\lim_{x \rightarrow \infty}\left(x-\sqrt{x^{2}+1}\right)\) can be rewritten as a fraction: \(\lim_{x \rightarrow \infty}\frac{x}{\frac{1}{\sqrt{x^2+1}}}\). After differentiating the numerator and denominator, the limit becomes \(\lim_{x \rightarrow \infty} \frac{(x^2 + 1)^{\frac{3}{2}}}{-x}\). Since \((x^2 + 1) ^{\frac{3}{2}}\) grows faster than -x as x approaches infinity, the limit is 0.

Step 1: Rewrite the expression as a fraction

First, let's rewrite the given expression in the form of a fraction to apply l'Hôpital's Rule: $$ \lim_{x \rightarrow \infty}\frac{x}{\frac{1}{\sqrt{x^2+1}}} $$

Step 2: l'Hôpital's Rule

Now we can apply l'Hôpital's Rule by differentiating numerators and denominators separately: $$ f(x) = x \Rightarrow f'(x) = 1 \\ g(x) = \frac{1}{\sqrt{x^2+1}} \Rightarrow g'(x) = \frac{-x}{(x^2+1)^{\frac{3}{2}}} $$ Then, we find the limit of the new expression: $$ \lim_{x \rightarrow \infty}\frac{1}{\frac{-x}{(x^2+1)^{\frac{3}{2}}}} $$

Step 3: Simplify and find the limit

To find the limit, we can simply divide the denominator by the numerator: $$ \lim_{x \rightarrow \infty} \frac{(x^2 + 1)^{\frac{3}{2}}}{-x} $$ The term \((x^2 + 1) ^{\frac{3}{2}}\) grows faster than -x as x approaches infinity. Therefore, the limit is: $$ \lim_{x \rightarrow \infty} \left(x-\sqrt{x^{2}+1}\right) = 0 $$

Key Concepts

Limits at infinity

When dealing with limits at infinity, we want to understand the behavior of a function as the variable approaches infinity, meaning as it gets larger and larger without bound. In calculus, this is crucial for describing the end behavior of functions.

Specifically, with the given exercise, we look at the expression \( x - \sqrt{x^2 + 1} \) as \( x \) approaches infinity. The challenge arises when we recognize that both terms \( x \) and \(\sqrt{x^2 + 1}\) are themselves becoming infinite, making it less clear what their difference approaches. This expression is an example of a difference of two large quantities, which could lead to an indeterminate form and requires special techniques, such as l'Hôpital's Rule, to evaluate.

Differentiation

In the context of solving limits, differentiation is a powerful tool. It involves finding the derivative of a function, which gives the rate at which the function's value changes with respect to a change in its input variable.

When applying l'Hôpital's Rule, as in the given exercise, we differentiate the numerator and denominator separately. It means if we have a function \( f(x) \) with the derivative \( f'(x) \) and a function \( g(x) \) with the derivative \( g'(x) \) in a fractional form \(\frac{f(x)}{g(x)}\), we can use their derivatives to find the new limit \(\frac{f'(x)}{g'(x)}\) if the original limit produces an indeterminate form. Proper differentiation is critical in ensuring that l'Hôpital's Rule is correctly applied.

Indeterminate Forms

Indeterminate forms pose a particular challenge in calculus. They occur when evaluating a limit results in an expression that cannot clearly be resolved into a specific value. Common forms include 0/0, \infty/\infty, 0\cdots\infty, \infty - \infty, 1^\infty, 0^0, and \infty^0.

These forms do not have an inherent value and hence require further manipulation to resolve. In our exercise, the subtraction \( x - \sqrt{x^2 + 1} \) approaches an \infty - \infty indeterminate form as \( x \) goes to infinity. By rewriting the expression and then using l'Hôpital's Rule, we can effectively evaluate these challenging limits. Recognizing indeterminate forms allows students to know when additional techniques, such as l'Hôpital's Rule, need to be applied to find a solution.

Calculus

The field of calculus is vast and encompasses a variety of concepts and techniques for analyzing and understanding change. It consists of two main branches: differential calculus, concerning rates of change and slopes of curves, and integral calculus, dealing with accumulation of quantities and areas under or between curves.

l'Hôpital's Rule, used in this exercise, is a component of differential calculus and is applied to resolve certain limits involving indeterminate forms. This rule exemplifies how calculus provides systematic approaches to problems that are otherwise not easily solved. Through the differentiation of functions and the evaluation of limits, calculus allows students to tackle complex mathematical concepts systematically and effectively.