Problem 53 - Calculus Single Variable

Question

In Exercises \(53-60,\) calculate the given limit. \(\lim _{x \rightarrow 0} \frac{\arctan (x)}{x}\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the form of the limit

As we evaluate the limit \( \lim _{x \rightarrow 0} \frac{\arctan(x)}{x} \), observe that both \( \arctan(x) \) and \( x \) approach 0 as \( x \rightarrow 0 \). This is an indeterminate form \( \frac{0}{0} \), which suggests applying a method like L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule allows us to differentiate the numerator and the denominator separately when dealing with a \( \frac{0}{0} \) form. Differentiate the numerator, \( \arctan(x) \), to get \( \frac{1}{1+x^2} \), and differentiate the denominator, \( x \), to get 1. Rewrite the limit as: \[\lim _{x \rightarrow 0} \frac{\arctan (x)}{x} = \lim _{x \rightarrow 0} \frac{\frac{1}{1+x^2}}{1}.\]

Step 3: Evaluate the simplified limit

Now that we have simplified the expression using L'Hôpital's Rule, evaluate the new limit: \[\lim _{x \rightarrow 0} \frac{1}{1 + x^2}.\]As \( x \rightarrow 0 \), the expression \( 1 + x^2 \) approaches 1, and hence the limit simplifies to \( \frac{1}{1} = 1 \).

Step 4: Conclusion

Since each application and simplification leads to the conclusion that effects due to \( x \) approaching 0 leave the constant terms unaffected, the entire limit evaluates to \( 1 \). Thus, the result of the limit is \( 1 \).

Key Concepts

Indeterminate Forms

Indeterminate forms occur in calculus when evaluating certain limits or expressions. They appear when a direct substitution of a limit leads to an undefined expression, like dividing zero by zero or multiplying infinity by zero.
These forms signal the need for additional techniques to determine the limit's actual value. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), and more.
Recognizing the indeterminate form helps decide which method, such as algebraic manipulation, limits algebra, or 'L’Hôpital’s Rule', to use. In this exercise, the limit \(\lim _{x \rightarrow 0} \frac{\arctan(x)}{x}\) qualified as an indeterminate form of \(\frac{0}{0}\).
By identifying it, we're informed that direct evaluation wouldn’t work, prompting the application of a different technique to precisely evaluate the limit.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for solving limits that appear as indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
It involves differentiating the numerator and the denominator separately and then evaluating the resulting limit. This rule simplifies otherwise complex limits, especially when direct substitution is insufficient.
In the given exercise, once the \(\frac{0}{0}\) form was identified for \(\lim _{x \rightarrow 0} \frac{\arctan(x)}{x}\), L'Hôpital's Rule was applied:

• Differentiate the numerator \(\arctan(x)\) to get \(\frac{1}{1+x^2}\).
• Differentiate the denominator \(x\) to get \(1\).

The limit was then re-evaluated as \(\lim _{x \rightarrow 0} \frac{\frac{1}{1+x^2}}{1}\), simplifying to \(\frac{1}{1} = 1\).
This result shows L'Hôpital's Rule transforming a complex form into a solvable expression.

Differentiation Techniques

Differentiation is a key concept in calculus used to find the rate at which a quantity changes. It forms the basis of understanding slopes and tangents of curves, and also plays a crucial role in determining limits through L'Hôpital's Rule.
To apply L'Hôpital's Rule, one must differentiate both the numerator and denominator of an expression individually. This uses common differentiation techniques:

• For the function \(\arctan(x)\), the derivative is calculated as \(\frac{1}{1+x^2}\).
• The derivative of \(x\) is simply \(1\).

This differentiation helps to convert a non-evaluable form into a simpler function that can be directly substituted with the limit value, as illustrated in evaluating the limit \(\lim _{x \rightarrow 0} \frac{\arctan(x)}{x}\).
Mastering differentiation can significantly ease solving complex calculus problems by enabling one to break down and analyze dynamic changes within mathematical functions.