/*! 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 39 Find the limit. $$\lim _{x \ri... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limit. $$\lim _{x \rightarrow 0} \frac{\sin 3 x}{x}$$

Short Answer

Expert verified
The limit is 3.

Step by step solution

01

Recognize the form that resembles a standard limit

The given limit expression \( \lim_{x \rightarrow 0} \frac{\sin 3x}{x} \) is similar to the standard limit \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \). To solve it, we can manipulate the expression to utilize this standard form.
02

Use substitution to simplify

Introduce a substitution by letting \( u = 3x \). Then as \( x \rightarrow 0 \), it follows that \( u \rightarrow 0 \) as well. We rewrite the limit in terms of \( u \): \[ \lim_{x \rightarrow 0} \frac{\sin 3x}{x} = \lim_{u \rightarrow 0} \frac{\sin u}{u/3} = 3 \lim_{u \rightarrow 0} \frac{\sin u}{u} \].
03

Apply the known limit

Now that the expression is in the form \( 3 \lim_{u \rightarrow 0} \frac{\sin u}{u} \), apply the standard limit result: \( \lim_{u \rightarrow 0} \frac{\sin u}{u} = 1 \). Substituting this result gives us: \( 3 \times 1 = 3 \).

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Key Concepts

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

Standard Limit
Limits are a fundamental concept in calculus, and one of the most useful standard limits is \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \). This limit is particularly helpful when dealing with expressions involving sine functions nearing zero. Understanding this limit is crucial because it forms the foundation for various calculations in calculus.
It is especially important for simplifying complex expressions in order to find an answer. For example, if you know this standard limit, you can easily handle scenarios where the variable inside the sine function multiplies by a constant, such as \( \sin(3x) \) in our exercise.
The standard limit emphasizes the behavior of sine near zero. Since \( \sin x \) and \( x \) approach zero at a similar rate, their ratio forms this predictable limit, simplifying many trigonometric limit challenges.
Substitution Method
The substitution method is a powerful tool for simplifying limit calculations, especially when they are not immediately in a standard form. By replacing the variable with another one, often called substitution variable, the expression becomes more manageable.
In our exercise, we made use of this by setting \( u = 3x \). This helps in transforming our original limit into a form that matches our known standard limits, making it easier to solve. Here are the steps involved with using substitution:
  • Identify the substitution that makes the expression simpler.
  • Change variables in the limit, ensuring you also change the variable approaching point.
  • Simplify the expression using known limits or algebraic manipulation.
By applying this method, complex trigonometric limits become easier to handle, enabling straightforward computation of the limit values.
Trigonometric Limits
Trigonometric limits form an essential part of calculus and often involve angles approaching zero or other significant angles. In calculus, understanding how trigonometric functions behave near particular points can unlock solutions to many problems involving curves, oscillations, and waves.
Common trigonometric limits often encountered include \( \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0 \). Knowing these can drastically reduce the complexity of limit problems involving trigonometric functions.
In our example, once the expression was reformed to \( \lim_{u \rightarrow 0} \frac{\sin u}{u} \), applying the known trigonometric limit directly brought us to the solution. Recognizing and memorizing such key limits helps in predicting the behavior of trig functions, which is a powerful strategy in solving limit exercises. This way, they become an invaluable resource when tackling anything from sine cosine expressions to more challenging trigonometric problems.

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Most popular questions from this chapter

On page 431 of Physics: Calculus, 2\(d\) ed, by Eugene Hecht (Pacific Grove, CA: Brooks/Cole, \(2000 ),\) in the course of deriving the formula \(\mathrm{T}=2 \pi \sqrt{\mathrm{L} / \mathrm{g}}\) for the period of a $$ \begin{array}{l}{\text { pendulum of length } L, \text { the author obtains the equation }} \\ {\mathrm{a}_{\mathrm{T}}=-g \sin \theta \text { for the tangential acceleration of the bob of the }}\end{array} $$ pendulum, He then says, "for small angles, the value of \(\theta\) in radians is very nearly the value of \(\sin \theta\) ; they differ by less than 2\(\%\) out to about \(20^{\circ} .\) $$ \begin{array}{c}{\text { (a) Verify the linear approximation at } 0 \text { for the sine function: }} \\ {\sin x=x}\end{array} $$ (b) Use a graphing device to determine the values of \(x\) for which sin \(x\) and \(x\) differ by less than 2\(\%\) . Then verify Hecht's statement by converting from radians to degrees.

Use the Chain Rule to prove the following. (a) The derivative of an even function is an odd function. (b) The derivative of an odd function is an even function.

Suppose \(y = f ( x )\) is a curve that always lies above the \(x\) -axis and never has a horizontal tangent, where \(f\) is differentiable everywhere. For what value of \(y\) is the rate of change of \(y ^ { 5 }\) with respect to \(x\) eighty times the rate of change of \(y\) with respect to \(x ?\)

Find the derivative. Simplify where possible. $$G(x)=\frac{1-\cosh x}{1+\cosh x}$$

\(23-28\) Use a linear approximation (or differentials) to estimate the given number. $$ 1 / 1002 $$

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