Problem 36 - Thomas Calculus

Question

Find the limits. Are the functions continuous at the point being approached? $$\lim _{x \rightarrow \pi / 6} \sqrt{\csc ^{2} x+5 \sqrt{3} \tan x}$$

Step-by-Step Solution

Answer

Limit is 3; the function is continuous at \(x = \frac{\pi}{6}\).

Step 1: Simplify the Expression Inside the Limit

First, we need to identify and possibly simplify the expression that is inside the limit: \[\sqrt{\csc ^{2} x+5 \sqrt{3} \tan x}\]Start by rewriting \(\csc x\) and \(\tan x\) in terms of \(\sin x\) and \(\cos x\):\(\csc x = \frac{1}{\sin x}\) and \(\tan x = \frac{\sin x}{\cos x}\). This gives us:\[\csc^2 x = \frac{1}{\sin^2 x}\]\[5\sqrt{3}\tan x = 5\sqrt{3}\frac{\sin x}{\cos x}\]

Step 2: Calculate the Limits of Each Component

Now, find the limits of each component as \(x \to \frac{\pi}{6}\).\(\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}\), thus \(\csc \left(\frac{\pi}{6}\right) = 2\) and \(\csc^2 \left(\frac{\pi}{6}\right) = 4\).\(\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\), hence \(5 \sqrt{3} \tan \left(\frac{\pi}{6}\right) = 5\).Substituting these values back gives:\[\lim _{x \to \frac{\pi}{6}} \csc^2 x + 5\sqrt{3} \tan x = 4 + 5 = 9\]

Step 3: Evaluate the Square Root and Find the Limit

Now, apply the square root to the limit of the sum we calculated:\[\sqrt{\lim _{x \to \frac{\pi}{6}} (\csc^2 x + 5\sqrt{3} \tan x)} = \sqrt{9} = 3\]Thus, the limit is \(3\).

Step 4: Determine Continuity at the Point

A function is continuous at a point if the limit as \(x\) approaches the point is equal to the function's value at that point. Here, we need \[x = \frac{\pi}{6}\]and compute:\[\sqrt{\csc^2 \left(\frac{\pi}{6}\right) + 5 \sqrt{3} \tan \left(\frac{\pi}{6}\right)} = \sqrt{4 + 5} = \sqrt{9} = 3\]Since the limit and the value at the point are both \(3\), the function is continuous at \(x = \frac{\pi}{6}\).

Key Concepts

Trigonometric Functions

Trigonometric functions are a cornerstone of calculus, particularly in evaluating limits involving angles. They are functions related to the angles of a triangle and include sine, cosine, tangent, cosecant, secant, and cotangent.

In the context of the limit given, we focus on the cosecant and tangent functions. Cosecant, \( \csc x \), is the reciprocal of the sine function, meaning \( \csc x = \frac{1}{\sin x} \). Tangent, \( \tan x \), is the ratio of the sine to the cosine: \( \tan x = \frac{\sin x}{\cos x} \).

This problem requires transforming these trigonometric identities to simplify complex expressions. Simplifying helps in finding the limit by expressing terms in forms easier to compute when evaluating limits.

Continuity

Continuity is a property of functions that describes the smoothness of their graphs. A function is continuous at a point if there are no breaks, jumps, or holes at that point.

In mathematical terms, a function \( f(x) \) is continuous at a point \( a \) if:

• The function \( f(x) \) is defined at \( x = a \).
• The limit of \( f(x) \) as \( x \to a \) exists.
• The limit of \( f(x) \) as \( x \to a \) equals \( f(a) \).

For the function given by the limit problem, we found that at \( x = \frac{\pi}{6} \), the limit is 3. Calculating the function directly at \( x = \frac{\pi}{6} \) also gives 3. Thus, these conditions of continuity are satisfied, indicating the function has no interruptions at this point.

Limit Evaluation

Limit evaluation is the process of determining the value that a function approaches as the input approaches a particular point. It's crucial for understanding the behavior of functions around that point.

To evaluate limits, especially when trigonometric identities are involved, simplifying the expression first is often necessary. This can be done by converting functions to their basic trigonometric forms, as seen in the original problem where \( \csc^2 x \) was converted to \( \frac{1}{\sin^2 x} \) and \( \tan x \) to \( \frac{\sin x}{\cos x} \).

Once simplified, we substitute the approaching point's value into the simplified expression. Here, at \( x = \frac{\pi}{6} \), relevant trigonometric values like \( \sin \frac{\pi}{6} = \frac{1}{2} \) and \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \) were used for evaluation. The calculated components, when added, provided the final limit. In this problem, the expression simplifies to a square root operation yielding the result \( 3 \). Understanding these steps is key to mastering limit evaluation in calculus.