Problem 66 - Calculus: Graphical, Numerical, Algebraic

Question

True or False \(\lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } = 2 .\) Justify your answer.

Step-by-Step Solution

Answer

True, \(\lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } = 2\).

Step 1: Recognize and simplify

Recognize that the limit is in the form where direct substitution would result in 0/0, an indeterminate form. Therefore, apply the property that \(\lim _ { x \rightarrow a } ( f(x) + g(x) ) = \lim _ { x \rightarrow a } f(x) + \lim _ { x \rightarrow a } g(x)\) to break it down as \(\lim _ { x \rightarrow 0 } \frac { x } { x } + \lim _ { x \rightarrow 0 } \frac { \sin x } { x }\)

Step 2: Evaluate the limit

As x approaches 0, the first part \(\lim _ { x \rightarrow 0 } \frac { x } { x }\) simplifies to 1 since any number divided by itself equals 1 (with exception of dividing by zero). The second part \(\lim _ { x \rightarrow 0 } \frac { \sin x } { x }\) is a standard limit that also equals 1. This is a well-known limit that comes from the definition of the derivative of sine. Therefore, we get 1 + 1 = 2.

Step 3: Interpret the result

The result tells us that, as x approaches 0, the quotient of the function \(x + \sin x\) and x approaches the limit 2, so the statement \(\lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } = 2\) is true.

Key Concepts

Indeterminate form

When dealing with limits, you might come across expressions that seem impossible to solve directly. Indeterminate forms are cases where you substitute the value into an expression and get an ambiguous form like \( \frac{0}{0} \). These forms do not readily provide any information about the limit.

In the expression given, \( \lim _ { x \rightarrow 0 } \frac { x + \sin x } { x } \), directly substituting \( x = 0 \) gives us \( \frac{0}{0} \), an indeterminate form. This tells us that we need to find another way to evaluate the limit.

One method to resolve this is to simplify or alter the expression. In many cases, breaking down the expression into simpler parts can help solve the indeterminacy. This process can employ algebraic manipulation, L'Hôpital's Rule, or by recognizing standard limits, which we'll discuss further in the next sections.

Standard limits

Standard limits are commonly known limits that are frequently used in calculus. They are derived from fundamental calculus concepts and can greatly simplify the process of finding limits of more complex functions.

These limits often come up in trigonometric functions and exponential functions. A popular example is the limit \( \lim _ { x \rightarrow 0 } \frac { \sin x } { x } \), which is equal to 1. This specific limit appears in trigonometry and is a standard identity used when working with derivatives and integration practices.

In our exercise, we separated the limit into two parts: \( \lim _ { x \rightarrow 0 } \frac { x } { x } \), which simplifies to 1, and \( \lim _ { x \rightarrow 0 } \frac { \sin x } { x } \), our standard limit which is also 1. By recognizing and applying these standard limits, the original expression simplifies to 1 + 1 = 2.

Trigonometric limits

Trigonometric limits are essential when dealing with functions involving sine, cosine, and other trigonometric operations. They help resolve indeterminate forms and simplify calculations in calculus.

One crucial trigonometric limit to remember is \( \lim _ { x \rightarrow 0 } \frac { \sin x } { x } = 1 \). This limit can be tested graphically by examining the graph of \( \frac { \sin x } { x } \), which approaches 1 as \( x \) approaches 0. It's also derived through limit theorems and serves as a foundational element in calculus.

By understanding this unique property, you can simplify complex limits involving trigonometric functions. In our problem, this knowledge allowed us to confidently state that \( \lim _ { x \rightarrow 0 } \frac { \sin x } { x } = 1 \), resulting in the final answer of 2 when added to the limit of \( \frac{x}{x} \). Knowing these fundamental trigonometric limits is incredibly helpful for quickly solving calculus problems.