91Ó°ÊÓ

L'Hôpital's Rule is a powerful theorem in calculus used to evaluate limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule states that if the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in one of these indeterminate forms, you can differentiate the numerator \( f(x) \) and the denominator \( g(x) \) separately and then take the limit of their quotient, \[ \lim_{x \to c} \frac{f'(x)}{g'(x)} \]. However, there are conditions to be met. First, both \( f \) and \( g \) must be differentiable near \( c \). Also, \( g'(x) eq 0 \) around \( c \), and the new limit must exist. L'Hôpital's Rule is repeatedly applied until a solvable form is reached, often simplifying otherwise tricky expressions.
In the given problem, the limit of \( \frac{(\tan^{-1} \sqrt{x})^2}{x \sqrt{x+1}} \) at \( x \rightarrow 0^{+} \) equals \( \frac{0}{0} \), allowing the application of L'Hôpital's Rule to differentiate and simplify the problem step by step.

Indeterminate forms occur when you attempt to calculate a limit and the function's forms are not definitive. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), and more.
These forms are labeled "indeterminate" because they don't readily lead to a clear answer. In such cases, techniques like L'Hôpital's Rule or series expansions are used to resolve them.
In our example, substituting \( x = 0^{+} \) in \( \frac{(\tan^{-1} \sqrt{x})^2}{x \sqrt{x+1}} \) produced the form \( \frac{0}{0} \). This signals the need for further analysis using differentiation or series expansion to find a solvable expression for the limit.

The Taylor Series is an approximation of a function as an infinite sum of terms calculated from the values of the function's derivatives at a single point. For a function \( f(x) \) about the point \( a \), the Taylor series is given by \[ T(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]. For limits, especially when facing indeterminate forms, it can simplify complex expressions by approximating functions to basic polynomial forms.In our exercise, the Taylor expansion of \( \tan^{-1} \sqrt{x} \) around zero simplifies the analysis by approximating \( \tan^{-1} \sqrt{x} \approx \sqrt{x} - \frac{x^{3/2}}{3} \). This turns the expression into a more manageable form, ultimately breaking down the limit near \( x=0 \) to find the solution.

Evaluating derivatives is a fundamental step in applying L'Hôpital's Rule. This involves finding the derivatives of the original function's numerator and denominator to form a new function whose limit is more readily evaluated. Differentiation techniques apply here, including chain rule, product rule, or quotient rule as needed.In our step-by-step solution:

• First, differentiate \( (\tan^{-1} \sqrt{x})^2 \) to get \( \frac{\tan^{-1} \sqrt{x}}{(1+x)\sqrt{x}} \)
• Next, differentiate \( x \sqrt{x+1} \) to obtain \( \frac{3x+2}{2\sqrt{x+1}} \)

These derivatives form a new limit expression using L'Hôpital's Rule again to tackle each part individually. It’s critical that the derivatives are taken correctly, ensuring calculations proceed smoothly and results align with expectations.