Problem 100 - Thomas Calculus

Question

Find the limits in Exercises \(97-104\) $$ \lim _{x \rightarrow 0} \frac{2 \tan ^{-1} 3 x^{2}}{7 x^{2}} $$

Step-by-Step Solution

Answer

The limit is \(\frac{6}{7}\).

Step 1: Identify the Form of the Limit

Firstly, substitute 0 into the expression to check if it results in an indeterminate form. \(\lim _{x \rightarrow 0} \frac{2 \tan ^{-1} 3 x^{2}}{7 x^{2}} = \frac{2 \tan ^{-1} 0}{7 \cdot 0} = \frac{0}{0}\) which indicates an indeterminate form, allowing the use of L'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator

Apply L'Hôpital's Rule, which involves differentiating the numerator and denominator separately:Numerator: \( f(x) = 2 \tan^{-1} (3x^2) \Rightarrow f'(x) = \frac{2 \cdot 6x}{1 + (3x^2)^2} = \frac{12x}{1 + 9x^4} \)Denominator: \( g(x) = 7x^2 \Rightarrow g'(x) = 14x \)

Step 3: Apply L'Hôpital's Rule

Use L'Hôpital's Rule:\[\lim _{x \rightarrow 0} \frac{\frac{12x}{1 + 9x^4}}{14x} = \lim _{x \rightarrow 0} \frac{12x}{14x(1 + 9x^4)}\]Simplify to:\[\lim _{x \rightarrow 0} \frac{12}{14(1 + 9x^4)} = \frac{12}{14} \cdot \lim _{x \rightarrow 0} \frac{1}{1 + 9x^4}\]

Step 4: Evaluate the Simplified Limit

Since \(\lim _{x \rightarrow 0} \frac{1}{1 + 9x^4} = 1\), the expression simplifies:\[\lim _{x \rightarrow 0} \frac{12}{14} \cdot 1 = \frac{6}{7}\]

Step 5: Conclusion

After applying L'Hôpital's Rule and simplifying, the limit is found to be \(\frac{6}{7}\).

Key Concepts

Indeterminate Forms

When dealing with limits, you'll occasionally encounter expressions that appear undefined or `indeterminate` at first glance. These often take the form of expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When this happens, it's known as an `indeterminate form`. This doesn't mean the limit doesn't exist; rather, it signals that further analysis is needed to find a finite result or a more definitive interpretation of the expression.

In this exercise, substituting \(x = 0\) into the given limit \(\lim_{x \rightarrow 0} \frac{2 \tan ^{-1} 3 x^{2}}{7 x^{2}} \) yields \(\frac{0}{0}\). This is why L'Hôpital's Rule can be applied, as it helps evaluate such forms by providing a method to differentiate both the numerator and the denominator until the indeterminate form is resolved.

Limits

`Limits` help us understand the behavior of functions as they approach specific points or infinity. They lay the groundwork for calculus and are used to precisely define other important concepts, such as continuity and derivatives.

In this context, the task is to evaluate the limit \(\lim_{x \rightarrow 0} \frac{2 \tan ^{-1} 3 x^{2}}{7 x^{2}} \). Initially, substituting \(x = 0\) results in an indeterminate form \(\frac{0}{0}\), signaling the next logical step: applying L'Hôpital's Rule. Finding the limit involves a step-by-step simplification process to calculate the exact behavior of the function as \(x\) approaches zero.

Differentiation

`Differentiation` is the process of finding the derivative of a function, which represents how the function changes as its input changes. In calculus, it plays a crucial role and is especially useful for finding the slope of the tangent line to a curve at a given point.

When applying L'Hôpital's Rule, differentiation helps to tackle indeterminate forms by separately differentiating the numerator and the denominator of the expression. In the original problem, the function in the numerator is \(f(x) = 2 \tan^{-1}(3x^2)\) and its derivative is found to be \(f'(x) = \frac{12x}{1 + 9x^4}\). Similarly, the denominator \(g(x) = 7x^2\) is differentiated to find \(g'(x) = 14x\). These derivatives help transform the original indeterminate form into a simpler expression.

Inverse Trigonometric Functions

`Inverse trigonometric functions` are the arcsine, arccosine, and arctangent functions, which help to find angles when given certain trigonometric values. For example, \(\tan^{-1}(x)\) gives the angle whose tangent is \(x\). These functions are vital in calculus when dealing with trigonometry-involved expressions.

In this problem, the function \(\tan^{-1}(3x^2)\) is present in the numerator. To differentiate it, we rely on the derivative of the inverse tangent function, which is \(\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \times \frac{du}{dx}\). Applying this formula aids in simplifying the expression until it no longer has an indeterminate form, leading to a more manageable computation of the limit.