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Chapter 3: Problem 39

Trigonometric Limit Evaluate: \(\lim _{x \rightarrow 0}\left(\frac{\sin 3 x}{5 x}\right)\)

Short Answer

Expert verified
The limit as \(x\) approaches \(0\) of the given function is \(\frac{3}{5}\).

Step by step solution

01

Recognize the indeterminate form

Firstly, substituting \(x=0\) in the function \(\frac{\sin 3x}{5x}\) gives \(0/0\), which is clearly an indeterminate form.
02

Apply L'Hopital's Rule

L'Hopital's Rule allows to evaluate this limit. According to the rule, if the limit of a function is in an indeterminate form of type \(0/0\) or \(\infty/\infty\), then the limit of the function can be found as the limit of the ratio of the derivatives of the numerator and the denominator. Hence, replace \(\frac{\sin 3x}{5x}\) with \(\frac{d/dx \sin3x}{d/dx 5x}\). The derivative of \(\sin 3x\) is \(3\cos3x\) and the derivative of \(5x\) is \(5\).
03

Evaluate the limit of the new function

After replacing the function with the ratio of the derivatives, we get \(\frac{3\cos3x}{5}\). Now, replace \(x\) with \(0\) and evaluate the limit. \(\cos3x\) at \(x=0\) turns out to be \(1\) and we get the limit as \(\frac{3}{5}\).

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Key Concepts

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

L'Hopital's Rule
L'Hopital's Rule is a powerful tool in calculus for solving limits that result in indeterminate forms such as \(0/0\) or \(\infty/\infty\). When a direct substitution into the limit leads to one of these forms, L'Hopital's Rule allows us to differentiate the numerator and the denominator separately and then re-evaluate the limit. This process can significantly simplify the limit expression.

Here's a simple way to apply L'Hopital's Rule:
  • Confirm that substituting the limit into your function yields an indeterminate form.
  • Apply differentiation to both the numerator and the denominator independently.
  • Re-evaluate the limit with the new expressions.
By transforming the expression through differentiation, L'Hopital's Rule converts the problematic limit into a more manageable form.
Indeterminate Forms
Indeterminate forms appear when direct substitution in a limit results in expressions like \(0/0\), \(\infty/\infty\), \(0 \times \infty\), and others. These forms do not provide enough information to determine the limit's value directly.

For example, when evaluating \( \lim _{x \rightarrow 0} \left( \frac{\sin 3x}{5x} \right) \): substituting \(x = 0\) gives \(\frac{0}{0}\), an indeterminate form.

A deeper understanding of indeterminate forms helps you recognize conditions where standard limit-solving techniques, like factorization or L'Hopital's Rule, are necessary. Learning to identify these forms aids significantly in correctly evaluating limits.
Derivatives of Trigonometric Functions
Understanding derivatives of trigonometric functions is crucial when applying L'Hopital's Rule to limits involving trigonometric expressions. Trigonometric derivatives are foundational in calculus. They transform trigonometric functions into simpler, more direct forms. Here are some key derivatives:
  • The derivative of \(\sin x\) is \(\cos x\).
  • The derivative of \(\cos x\) is \(-\sin x\).
  • The derivative of \(\tan x\) is \(\sec^2 x\).


In the original limit problem, the derivative of \(\sin 3x\) was needed. We observed that it is \(3\cos 3x\), a result of applying the chain rule. Knowing these derivatives allows us to navigate through limits and apply L'Hopital's Rule smoothly and accurately.

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Most popular questions from this chapter

Evaluate: \(\lim _{x \rightarrow 0}\left(\frac{\left(1+3 x+2 x^{2}\right)^{1 / x}-\left(1+3 x-2 x^{2}\right)^{1 / x}}{x}\right)\)

The value of \(\lim _{n \rightarrow \infty}\left(\frac{a-1+\sqrt[n]{b}}{a}\right)^{n}, n \in N\) is (a) \(\sqrt[4]{b}\) (b) \(\sqrt[b]{a}\) (c) \(\sqrt{b}\) (d) \(\sqrt{a}\).

The value of \(\lim _{x \rightarrow 1}\left(\tan \left(\frac{\pi}{4}+\log x\right)\right)^{\frac{1}{\log x}}\) is (a) \(e\) (b) \(1 / e\) (c) \(e^{2}\) (d) \(1 / 2 e\)

Evaluate: \(\lim _{x \rightarrow \pi^{\prime}}\left(\frac{2^{\cot x}+3^{\operatorname{coc} x}-5^{1+\cot x}+10}{\left(4^{\cot x}\right)^{1 / 2}+\left(27^{\cot x}\right)^{1 / 3}-5^{\cot x}+20}\right)\)

L'Hospital Rule Evaluate: \(\lim _{x \rightarrow 0}\left(\frac{e^{x^{3}}-1-x^{3}}{64 x^{6}}\right)\)
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