Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each limit. $$ \lim _{x \rightarrow 0} \frac{\cos x}{x+1} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Substitute the Limit Value

First, we substitute the value that \(x\) approaches in the limit expression. So, substitute \(x = 0\) into the expression \(\frac{\cos x}{x+1}\). This gives us \(\frac{\cos 0}{0+1}\).

Step 2: Evaluate the Expression after Substitution

Now, calculate the expression \(\frac{\cos 0}{0+1}\). We know from trigonometry that \(\cos 0 = 1\). Therefore, the expression simplifies to \(\frac{1}{1}\).

Step 3: Simplify the Result

Simplify the expression \(\frac{1}{1}\) to find the value of the limit. This simplifies directly to \(1\).

Key Concepts

Evaluating Limits

Evaluating limits is a method in calculus used to determine the value that a function approaches as a variable gets closer to a particular point. Understanding how to evaluate limits is vital for solving problems involving continuous functions.
To evaluate a limit, you start by analyzing the function as the variable within it approaches the desired value. Evaluating limits can involve direct substitution, simplification, or the application of limit laws.
Here are some essential tips for evaluating limits:

• Direct Substitution: First try plugging the value the variable is approaching into the function, assuming the function is continuous at that point.
• Algebraic Simplification: If direct substitution leads to an indeterminate form, try to simplify the function algebraically.
• Understanding Indeterminate Forms: These occur often in limits, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). In these cases, more advanced techniques like the L'Hopital's Rule might be needed.

For the problem \(\lim _{x \rightarrow 0} \frac{\cos x}{x+1}\), we start with direct substitution since the function is continuous at 0.

Substitution in Limits

Substitution in limits is one of the simplest and most straightforward methods to evaluate a limit. The idea is to directly replace the variable in the limit expression with the value it approaches.
When the function is continuous and well-behaved at that particular point, substitution provides an immediate answer without further steps. However, in many scenarios where substitution does not work initially, it can be used in conjunction with simplifying the expression or applying other calculus techniques.
In our example problem, we use substitution by plugging \(x = 0\) directly into the expression \(\frac{\cos x}{x+1}\). This makes our expression evaluate to \(\frac{\cos 0}{0+1}\). Knowing that \(\cos(0) = 1\), it becomes \(\frac{1}{1}\), a straightforward result of 1.

Trigonometric Limits

Trigonometric limits often arise in calculus, especially when dealing with functions involving sine, cosine, and tangent. These limits are essential in various fields like physics and engineering to model wave behavior and oscillations.
Understanding trigonometric values at key angles, such as 0, \pi/2\, \pi\, and their multiples, helps in evaluating these limits.
For our specific problem, \(\lim _{x \rightarrow 0} \frac{\cos x}{x+1}\), we utilized the well-known fact that \(\cos(0)=1\). This knowledge is vital when solving trigonometric limits, as it allows us to approach problems with more confidence.

• Trigonometric Limits often involve special angle knowledge as well as trigonometric identities which simplify complex expressions.
• Remember that trigonometric function properties, such as even and odd functions, periodicity, and limits of trigonometric functions as they approach specific values, play a crucial role.

Hence, mastering these fundamental concepts can make you adept at solving various calculus problems involving limits.