Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x} $$

Short answer

The limit is 1.

Step-by-step solution

Identify the Form of the Limit

Examine the limit \( \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x} \). As \( x \rightarrow 0 \), both the numerator and the denominator approach zero, giving an indeterminate form \( \frac{0}{0} \). This suggests we can apply L'Hôpital's Rule.

Apply L'Hôpital's Rule

Since the limit is in an indeterminate form, apply L'Hôpital's Rule. This involves differentiating the numerator and the denominator separately. - Differentiate the numerator: \( \frac{d}{dx}(\sqrt{1+x} - \sqrt{1-x}) = \frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}} \).- Differentiate the denominator: \( \frac{d}{dx}(x) = 1 \).Thus, the limit becomes \( \lim _{x \rightarrow 0} \frac{\frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}}}{1} \).

Evaluate the Limit After Differentiation

Now, substitute \( x = 0 \) into the differentiated limit:\[ \lim _{x \rightarrow 0} \left( \frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}} \right) = \frac{1}{2\sqrt{1+0}} + \frac{1}{2\sqrt{1-0}} = \frac{1}{2} + \frac{1}{2} = 1. \]

Confirm the Result

After using L'Hôpital's Rule and evaluating the limit, we find that the initial limit \( \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x} = 1 \). Verify that the computed derivatives and substitutions were handled correctly to ensure no mistakes were made.

Key concepts

Indeterminate Forms

When analyzing limits in calculus, you may encounter expressions that initially seem undefined. These are known as indeterminate forms. They occur when a function's numerator and denominator both approach a value that makes the fraction appear undefined, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In these cases, straightforward substitution does not provide a clear answer. Instead, these suggest using advanced techniques like L'Hôpital's Rule to resolve them.

In the given exercise, we examined \( \lim_{x \to 0} \frac{\sqrt{1+x}-\sqrt{1-x}}{x} \). Both the numerator and the denominator approach zero as \( x \) approaches 0. When both the top and bottom of a fraction tend towards zero, it indicates the form \( \frac{0}{0} \), making it an ideal candidate for L'Hôpital's Rule. This rule helps solve such problems by differentiating the numerator and the denominator to find the limit.

Differentiation

Differentiation is the process of finding the derivative, or rate of change, of a function. It is a core concept in calculus that allows you to analyze how functions change over small intervals. This concept is crucial when applying L'Hôpital's Rule, as it requires finding derivatives of both the numerator and denominator of a limit expression.

In the context of our exercise, we need to differentiate \( \sqrt{1+x} - \sqrt{1-x} \) and \( x \). First, differentiate \( \sqrt{1+x} \), which results in \( \frac{1}{2\sqrt{1+x}} \), using the chain rule. A similar approach works for \( \sqrt{1-x} \), resulting in another \( \frac{1}{2\sqrt{1-x}} \). Differentiating \( x \) is straightforward, yielding 1. By finding these derivatives, we transform our problem into an easier limit evaluation, which is often much more manageable.

Limit Evaluation

Evaluating limits is a fundamental concept in calculus that involves finding the value a function approaches as the input approaches a certain point. When direct substitution in a limit leads to an indeterminate form, techniques like L'Hôpital’s Rule become invaluable.

Once you've applied differentiation to handle the indeterminate form \( \frac{0}{0} \), the next step is substituting back into the limit expression. In this exercise, after differentiating both parts of the fraction, we substitute \( x = 0 \) back into the differentiated forms: \( \frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}} \). Each half becomes \( \frac{1}{2} \) as \( x \rightarrow 0 \) simplifies into \( \frac{1}{2\sqrt{1}} \). Therefore, the limit evaluates neatly to 1.

This sequence of steps not only simplifies the initially complex expression but also ensures accuracy, confirming the solution aligns with mathematical principles.