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Chapter 1: Problem 32

Find the limits. $$\lim _{x \rightarrow 0} \frac{\sin ^{2}(k x)}{x}, \quad k \neq 0$$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Recognize Indeterminate Form

First, recognize that as \(x\) approaches 0, the expression \(\frac{\sin^2(kx)}{x}\) is in the indeterminate form \(\frac{0}{0}\). Therefore, we need to apply some method to simplify and evaluate the limit.
02

Apply Limit Substitution

Consider using the fact that \(\lim_{y \to 0} \frac{\sin y}{y} = 1\). For our problem, let \(y = kx\), then as \(x\) approaches 0, \(y\) also approaches 0.
03

Algebraic Manipulation

Rewrite the expression using the substitution: \(\frac{\sin^2(kx)}{x} = \frac{\sin(kx) \cdot \sin(kx)}{x} = \frac{\sin(kx)}{kx} \cdot k \cdot \sin(kx)\).
04

Apply Limit Fact

Using the limit fact \(\lim_{y \to 0} \frac{\sin y}{y} = 1\), we substitute \(y = kx\) to get \(\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1\). Therefore, the expression becomes \(k \cdot \lim_{x \to 0} \sin(kx) \cdot 1 = k \cdot 0 = 0\).
05

Final Limit Evaluation

Conclude that the entire expression equals 0 due to the multiplication by sin(kx), which approaches 0 as \(x\) approaches 0. Thus, the limit \(\lim _{x \rightarrow 0} \frac{\sin ^{2}(k x)}{x} = 0.\)Therefore, the limit of the original expression as \(x\) approaches 0 is 0.

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Key Concepts

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

Indeterminate Forms
In calculus, limits can sometimes lead to indeterminate forms, such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), or \(0 \cdot \infty\). These forms occur when direct substitution in the limit results in undefined expressions.
Understanding indeterminate forms is crucial as they indicate that a more strategic approach is needed to evaluate the limit.
For example, in the problem \(\lim_{x \rightarrow 0} \frac{\sin^2(kx)}{x}\), both the numerator and the denominator approach zero as \(x\) approaches zero.
This results in the indeterminate form \(\frac{0}{0}\), prompting the use of algebraic manipulation or limit theorems to resolve the expression and find a definitive limit.
Limit Substitution
Limit substitution is a powerful technique used to simplify complicated-looking limits. This involves substituting a part of the original expression with a simpler variable.
In our example problem, the substitution \(y = kx\) can be used.
As \(x\) approaches 0, the variable \(y\) also approaches 0, transforming the limit problem into a simpler one.
This switch allows us to apply known limit results, such as \(\lim_{y \to 0} \frac{\sin y}{y} = 1\), making the evaluation easier.
  • Reduces complexity by breaking down functions into simpler parts
  • Enhances understanding by substituting with familiar forms
  • Essential for tackling indeterminate forms effectively
Sine Function Limits
The sine function has specific properties that are useful in solving limits, particularly the standard result \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).
This property is very handy, especially when dealing with limits involving trigonometric functions.
In the original exercise, the expression required adjusting into a form where this property applies.
By making a substitution and then recognizing the standard form \(\frac{\sin(kx)}{kx}\), the complex limit transformed into a simple multiplication problem.
  • It serves as a foundation for solving many similar trigonometric limits.
  • Facilitates simplification of complex expressions by aligning with basic sine limits.
  • Enables effective resolution of indeterminate forms by recognizing standard patterns.

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Most popular questions from this chapter

Find the limits. $$\lim _{x \rightarrow+\infty} \ln \left(\frac{2}{x^{2}}\right)$$

The notion of an asymptote can be extended to include curves as well as lines. Specifically, we say that curves \(y=f(x)\) and \(y=g(x)\) are asymptotic as \(x \rightarrow+\infty\) provided $$ \lim _{x \rightarrow+\infty}[f(x)-g(x)]=0 $$ and are asymptotic as \(x \rightarrow-\infty\) provided $$ \lim _{x \rightarrow-\infty}[f(x)-g(x)]=0 $$ In these exercises, determine a simpler function \(g(x)\) such that \(y=f(x)\) is asymptotic to \(y=g(x)\) as \(x \rightarrow+\infty\) or \(x \rightarrow-\infty\) Use a graphing utility to generate the graphs of \(y=f(x)\) and \(y=g(x)\) and identify all vertical asymptotes. $$f(x)=\frac{x^{5}-x^{3}+3}{x^{2}-1}$$

(a) Find the smallest positive number \(N\) such that for each \(x\) in the interval \((N,+\infty),\) the value of the function \(f(x)=1 / x^{2}\) is within 0.1 unit of \(L=0\) (b) Find the smallest positive number \(N\) such that for each \(x\) in the interval \((N,+\infty),\) the value of \(f(x)=\) \(x /(x+1)\) is within 0.01 unit of \(L=1\) (c) Find the largest negative number \(N\) such that for each \(x\) in the interval \((-\infty, N)\), the value of the function \(f(x)=1 / x^{3}\) is within 0.001 unit of \(L=0\) (d) Find the largest negative number \(N\) such that for each \(x\) in the interval \((-\infty, N),\) the value of the function \(f(x)=x /(x+1)\) is within 0.01 unit of \(L=1\)

Suppose that \(f\) is an invertible function, \(f(0)=0, f\) is continuous at \(0,\) and \(\lim _{x \rightarrow 0} f(x) / x\) exists. Given that \(L=\lim _{x \rightarrow 0} f(x) / x,\) show $$\lim _{x \rightarrow 0} \frac{x}{f^{-1}(x)}=L$$ [Hint: Apply Theorem 1.5 .5 to the composition hog, where $$h(x)=\left\\{\begin{array}{ll} f(x) / x, & x \neq 0 \\ L, & x=0 \end{array}\right.$$ and \(\left.g(x)=f^{-1}(x)\right].\)

Evaluate the limit using an appropriate substitution. (See Exercises \(45-46\) of Section \(1.3 .\) ) $$\lim _{x \rightarrow+\infty} \frac{\ln 2 x}{\ln 3 x}[\text {Hint: } t=\ln x]$$
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