91Ó°ÊÓ

Limit evaluation is a fundamental concept in calculus and is used to determine the value that a function approaches as the input approaches a certain point. It involves analyzing functions as they get indefinitely close to a particular input value.
In the original exercise, we look at the problem \[\lim _{x \rightarrow 0} \frac{\sin ^{-1} 5 x}{x} \]To solve this, we recognize it as evaluating a specific type of limit problem often encountered around indeterminate forms or functions involving trigonometry.
There are several techniques to evaluate limits:

• Direct substitution: Plugging the value into the function directly, when applicable.
• Factoring: Simplifying complex expressions by factoring to cancel terms.
• L'Hospital's Rule: Used when limits result in indeterminate forms.

In this problem, since direct substitution and factoring do not resolve the limit directly, L'Hospital's Rule becomes an effective method to evaluate it.

Indeterminate forms are expressions in calculus where a direct limit computation doesn't yield a clear answer. When attempting to evaluate limits, it's common to encounter indeterminate forms like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\) among others.
In our exercise, as \(x\) approaches 0, both the numerator \(\sin^{-1}(5x)\) and the denominator \(x\) approach 0, creating a \(\frac{0}{0}\) indeterminate form.
This is where L'Hospital's Rule can be particularly useful. It allows us to handle these indeterminate forms by taking derivatives:

• Differentiate the numerator: The derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Applying the chain rule for \(\sin^{-1}(5x)\) gives us: \(\frac{5}{\sqrt{1-(5x)^2}}\).
• Differentiate the denominator: The derivative of \(x\) is 1.

After differentiating, the limit becomes a straightforward evaluation, akin to what one might solve with basic algebraic limits.

Inverse trigonometric functions, such as \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\), are the inverse operations of the basic trigonometric functions. They allow us to find angles based on trigonometric values. For instance, \(\sin^{-1}(x)\) represents the angle whose sine is \(x\).
The derivative of inverse trigonometric functions is critical in calculus, especially when evaluating limits involving these functions.
For the derivative of \(\sin^{-1}(x)\), it's given by:\[\frac{1}{\sqrt{1-x^2}}\]Applying chain rule, specifically with \(\sin^{-1}(5x)\):

• The chain rule states: If \(y = f(g(x))\), then \(y' = f'(g(x)) \cdot g'(x)\).
• Here, \(g(x) = 5x\), which derives to 5, then multiply by the derivative of \(\sin^{-1}(u)\) where \(u=5x\).
• This yields \(\frac{5}{\sqrt{1-(5x)^2}}\).

Understanding these derivatives not only assists in limit evaluations but also helps in areas such as optimization and solving integrals.