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Find the limit, if it exists. $$ \lim _{x \rightarrow(\pi / 2)^{-}} \frac{\tan x}{\cot 2 x} $$

Short Answer

Expert verified
The limit exists and is -2.

Step by step solution

01

Understand the Problem

We need to find the limit of the expression \( \frac{\tan x}{\cot 2x} \) as \( x \) approaches \( \frac{\pi}{2} \) from the left. \( \tan x \) and \( \cot x \) are trigonometric functions that involve sine and cosine.
02

Simplify the Expression

Recall that \( \tan x = \frac{\sin x}{\cos x} \) and \( \cot 2x = \frac{\cos 2x}{\sin 2x} \). Therefore, the expression becomes \( \frac{\sin x \cdot \sin 2x}{\cos x \cdot \cos 2x} \).
03

Use Trigonometric Identities

The identity \( \sin 2x = 2 \sin x \cos x \) can simplify the expression further. Substitute it in to get: \( \frac{\sin x \cdot 2 \sin x \cos x}{\cos x \cdot \cos 2x} = \frac{2 \sin^2 x}{\cos 2x} \).
04

Evaluate the Expression Near the Limit

As \( x \rightarrow \frac{\pi}{2}^{-} \), \( \sin x \rightarrow 1 \), \( \cos x \rightarrow 0 \), and \( \cos 2x \rightarrow -1 \) since \( 2x \rightarrow \pi \). Simplifying each component: \( 2 \sin^2 x \rightarrow 2 \cdot 1^2 = 2 \) and \( \cos 2x \rightarrow -1 \). So, the limit becomes \( \frac{2}{-1} = -2 \).
05

Conclude the Solution

The calculations show that as \( x \) approaches \( \frac{\pi}{2} \) from the left-hand side, the limit of the given expression simplifies correctly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Trigonometric identities are powerful tools that allow us to simplify expressions involving trigonometric functions such as sine, cosine, tangent, and cotangent. Understanding and using these identities can make complex expressions much easier to work with.

In this exercise, we used the identities:
  • \( \tan x = \frac{\sin x}{\cos x} \)
  • \( \cot x = \frac{\cos x}{\sin x} \)
  • \( \sin 2x = 2\sin x \cos x \)
Using these identities, we transformed the expression \( \frac{\tan x}{\cot 2x} \) into a simpler form: \( \frac{2 \sin^2 x}{\cos 2x} \). This step is crucial because it simplifies the evaluation of limits by expressing the trigonometric functions in terms of more familiar operations.

Always remember, simplifying with identities not only aids in calculations but also gives us deeper insights into relationships between trigonometric functions.
Limit Evaluation
Limit evaluation is a method used to find the value that a function approaches as the input approaches some value. This is especially crucial in calculus, as it helps us understand the behavior of functions at points of discontinuity or where they become undefined.

In the given exercise, we are tasked with finding \( \lim_{x \rightarrow (\pi / 2)^{-}} \frac{2 \sin^2 x}{\cos 2x} \). As \( x \) approaches \( \frac{\pi}{2} \) from the left, we observe:
  • \( \sin x \rightarrow 1 \)
  • \( \cos x \rightarrow 0 \)
  • \( \cos 2x \rightarrow -1 \) because \( 2x \rightarrow \pi \)
By substituting these values into the simplified expression, the limit evaluates to \( \frac{2 \cdot 1^2}{-1} = -2 \).

This process shows how using the simplified form and understanding the behavior of trigonometric functions near specific points help us to evaluate limits effectively.
Continuity and Limits
Understanding continuity and limits is integral to calculus. A function is continuous at a point if it doesn't have breaks, jumps, or holes at that point. Limits describe what a function approaches as the input nears a particular value, whether or not the function is actually defined at that point.

In this exercise, we approach \( \frac{\pi}{2} \) from the left, denoted \( (\pi/2)^{-} \). At this point, \( \tan x \) and \( \cot 2x \) are not defined due to division by zero by their definitions. However, by using limits, we can predict the behavior of the function near \( \frac{\pi}{2} \).

Additionally, when we reduce expressions using trigonometric identities, we often remove potential discontinuities, making the limit evaluations at boundary points less complex. Thus, the concept of limits helps us navigate through points where direct evaluation might fail, ensuring a deeper understanding of function behavior.

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Most popular questions from this chapter

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