/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Evaluate the following limits. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following limits. $$\lim _{x \rightarrow 0} \csc 6 x \sin 7 x$$

Short Answer

Expert verified
The value of the limit is $$\frac{7}{6}$$.

Step by step solution

01

1. Rewrite the expression in terms of sine and cosine functions

We can rewrite the given expression as follows: $$ \lim_{x \rightarrow 0} \frac{\sin 7x}{\sin 6x} = \lim_{x \rightarrow 0} \frac{\sin (6x+x)}{\sin 6x} $$
02

2. Simplify the expression using the trigonometric identity \(\sin (A + B)\)

Using the trigonometric identity, we have: $$ \lim_{x \rightarrow 0} \frac{\sin (6x) \cos x + \cos (6x) \sin x}{\sin 6x} $$
03

3. Separate the expression into two fractions

Divide each term in the numerator by \(\sin 6x\): $$ \lim_{x \rightarrow 0} \left[\cos x \left(\frac{\sin(6x)}{\sin(6x)}\right) + \frac{\cos(6x)\sin x}{\sin(6x)} \right] $$
04

4. Simplify the first term

Since \(\frac{\sin(6x)}{\sin(6x)} = 1\), we can simplify the expression as: $$ \lim_{x \rightarrow 0} [\cos x + \frac{\cos (6x) \sin x}{\sin (6x)}] $$
05

5. Subdivide the limit

Compute the limit of each term separately: $$ \lim_{x \rightarrow 0} \cos x + \lim_{x \rightarrow 0} \frac{\cos (6x) \sin x}{\sin (6x)} $$
06

6. Evaluate the first limit

The first limit is very straightforward: $$ \lim_{x \rightarrow 0} \cos x = \cos 0 = 1 $$
07

7. Apply L'Hôpital's Rule to the second limit

We have an indeterminate form of the type \(\frac{0}{0}\), so we can apply L'Hôpital's Rule by taking the derivative of both the numerator and the denominator with respect to \(x\): $$ \lim_{x \rightarrow 0} \frac{d(\cos (6x) \sin x)}{d(\sin (6x))} = \lim_{x \rightarrow 0} \frac{-6\sin (6x) \sin x + \cos (6x) \cos x}{6\cos (6x)} $$
08

8. Evaluate the limit

As \(x \rightarrow 0\), we have: $$ \lim_{x \rightarrow 0} \frac{-6\sin (6x) \sin x + \cos (6x) \cos x}{6\cos (6x)} = \frac{-6\sin (0) \sin 0 + \cos (0) \cos 0}{6\cos (0)} = \frac{1}{6} $$
09

9. Add both limits together

Finally, add both individual limits together to find the limit of the given expression: $$ \lim _{x \rightarrow 0} \csc 6 x \sin 7 x = 1 + \frac{1}{6} = \frac{7}{6} $$

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Key Concepts

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

Limits in Calculus
Limits form the bedrock of calculus, allowing us to describe the behavior of functions as they approach specific points or infinity. They capture the essence of function trends, even when the function's value is undefined at the point in question.

For example, imagine you're examining a function as its input gets closer and closer to zero. The limit tells us what the output of the function approaches during this process. In mathematical terms, we denote this using the notation \( \lim_{x \rightarrow a} f(x) = L \) where \(a\) is the point we're approaching and \(L\) is the value that \(f(x)\) is getting closer to, or the limit. In essence, limits allow us to investigate the immediate neighborhood around a point, even if that point itself is not part of the function's domain.

Understanding limits is crucial when dealing with continuous functions, as it supports fundamental concepts like the derivative, which is, in effect, a limit of the ratio of changes in the function's output and input.
L'Hôpital's Rule
L'Hôpital's Rule is a fantastic tool for calculating limits of indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with such a form, instead of hitting the breaks, we dive into calculus and evaluate the limit by taking derivatives.

The rule states that if you have a limit of the form \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) and both \(f(x)\) and \(g(x)\) approach 0 or both approach infinity as \(x\) approaches \(a\), then this limit can be evaluated as \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \) provided this new limit exists or equals infinity.

By applying L'Hôpital's Rule to the problem at hand, one can often simplify the process of finding limits that would otherwise be challenging to solve. The rule's practicality is illustrated in the step-by-step solution, where it is used to resolve the indeterminate state resulting from the sine functions approaching zero.
Trigonometric Limits
Limits involving trigonometric functions are common in calculus, and they have their own set of rules and identities that can substantially simplify calculations.

A classic example is provided by the limit \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \) which is the foundation for many trigonometric limits. To calculate more complex trigonometric limits, like the one in our exercise, we can employ trigonometric identities, such as \( \sin (A + B) \) to break down the functions into simpler parts that are easier to handle.

These simplifications let us evaluate the limit in pieces, often allowing us to bypass indeterminate forms by recognizing patterns that lead to definitive values. The trigonometric identities not only assist in finding trigonometric limits but are also helpful when integrating functions or solving differential equations.
Indeterminate Forms
Indeterminate forms represent situations where limits lack evident outcomes and may imply multiple or conflicting results. These can occur in various disguises, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \).

These forms don't immediately suggest a specific limit, so alternative methods must be employed to find a definitive answer. Techniques used to resolve these ambiguities include factoring, conjugation, or applying L'Hôpital's Rule as seen in the exercise's step-by-step solution. The skill in dealing with indeterminate forms lies in transforming them into a state that allows the limit to be evaluated, revealing that behind an indeterminate veil often lies a tangible and calculable limit.

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