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91Ó°ÊÓ

Find the limits in Exercises 21–36. $$ \lim _{t \rightarrow 0} \frac{\sin (1-\cos t)}{1-\cos t} $$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Identify the Form of the Limit

We need to evaluate \( \lim_{t \rightarrow 0} \frac{\sin(1-\cos t)}{1-\cos t} \). Observing the expression, as \( t \rightarrow 0 \), both the numerator and the denominator go to zero, forming a \( \frac{0}{0} \) indeterminate limit.
02

Expand Using Taylor Series

To resolve the indeterminate form, use the Taylor series expansion around \( t = 0 \). The expansion is \( 1-\cos t = \frac{t^2}{2} - \frac{t^4}{24} + \cdots \). For \( \sin(1-\cos t) \), substitute \( 1-\cos t \) in the Taylor expansion of \( \sin x \), which is \( x - \frac{x^3}{6} + \cdots \).
03

Simplify Numerator and Denominator

Using the substitution from Step 2, \( \sin(1-\cos t) \approx \left( \frac{t^2}{2} \right) - \frac{1}{6}\left( \frac{t^2}{2} \right)^3 + \cdots \approx \frac{t^2}{2} \). Similarly, \( 1-\cos t \approx \frac{t^2}{2} \).
04

Evaluate the Simplified Limit

With both the numerator and denominator approximating to \( \frac{t^2}{2} \), the expression becomes \( \frac{\frac{t^2}{2}}{\frac{t^2}{2}} \), which simplifies to \( 1 \).
05

Finalize the Limit Solution

After simplifying, the expression resolves to a constant as \( t \rightarrow 0 \). Therefore, the limit is \( 1 \).

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Key Concepts

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

Indeterminate Forms
In Calculus, indeterminate forms often arise in the context of limits. These forms indicate expressions where direct substitution does not immediately yield a meaningful value. Instead, they appear as undefined forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 0 \cdot \infty \). In the original exercise, we encounter an indeterminate form \( \frac{0}{0} \).Recognizing an indeterminate form is the first step towards solving complex limit problems. It tells us that we need other techniques, like algebraic manipulation, L'Hôpital's rule, or Taylor series expansions, to find the limit. In our case, since direct substitution led to an indeterminate form, we opted for a Taylor series expansion to simplify and resolve the limit effectively.
Taylor Series Expansion
The Taylor series is a powerful tool used to approximate functions around a specific point, usually zero. In our example, we expanded both the numerator and the denominator using their respective Taylor series. For \( 1 - \cos t \), we used:
  • \( 1 - \cos t = \frac{t^2}{2} - \frac{t^4}{24} + \cdots \)
This series gives us the behavior of \( \cos t \) near zero.Next, we approximated \( \sin(1 - \cos t) \) through a substitution in the Taylor series for \( \sin x \):
  • \( \sin x = x - \frac{x^3}{6} + \cdots \)
Substituting \( x = 1 - \cos t \), we simplified \( \sin(1 - \cos t) \approx \frac{t^2}{2} \), neglecting higher order terms.Using these approximations, we simplified the expression \( \frac{\sin(1 - \cos t)}{1 - \cos t} \) by cancelling out equal terms in the numerator and denominator. This makes Taylor series expansions particularly useful in resolving limits presented in indeterminate forms.
Trigonometric Limits
Trigonometric limits involve functions like \( \sin t \) or \( \cos t \) as the variable approaches a certain point, often zero. Essential trigonometric limits frequently appear in calculus as these functions have behaviors specific to their domains. For example, understanding limits like \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) are fundamental building blocks.In our problem, we analyzed the behavior of \( \sin(1 - \cos t) \) as \( t \rightarrow 0 \). By applying Taylor series expansions to rearrange and simplify, we managed to reduce the expression to a form where normal limit techniques could be applied. This approach is beneficial because it allows us to
  • Determine limits without unsure direct substitution
  • Avoid complications that arise from small-angle approximations in trigonometric functions
By focusing on series expansions of trigonometric functions around zero, we can tackle challenging limit problems with more confidence.

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Most popular questions from this chapter

In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ h(t)=t^{3}, \quad(2,8) $$

In Exercises \(11-18,\) find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$ f(x)=x^{2}+1, \quad(2,5) $$

In Exercises \(19-22,\) find the slope of the curve at the point indicated. $$ y=1-x^{2}, \quad x=2 $$

Does the graph of $$ U(x)=\left\\{\begin{array}{ll}{0,} & {x<0} \\ {1,} & {x \geq 0}\end{array}\right. $$ have a vertical tangent at the point \((0,1) ?\) Give reasons for your answer.

In Exercises \(61-66,\) you will further explore finding deltas graphically. Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(x_{0}\) being approached. b. Guess the value of the limit \(L\) and then evaluate the limit symbolically to see if you guessed correctly. c. Using the value \(\epsilon=0.2,\) graph the banding lines \(y_{1}=L-\epsilon\) and \(y_{2}=L+\epsilon\) together with the function \(f\) near \(x_{0}\) . d. From your graph in part (c), estimate a \(\delta > 0\) such that for all \(x\) $$ 0 < \left|x-x_{0}\right| < \delta \quad \Rightarrow \quad|f(x)-L| < \epsilon $$ Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0 < \left|x-x_{0}\right| < \delta .\) For your viewing window use \(x_{0}-2 \delta \leq x \leq x_{0}+2 \delta\) and \(L-2 \epsilon \leq y \leq L+2 \epsilon\) . If any function values lie outside the interval \([L-\epsilon, L+\epsilon],\) your choice of \(\delta\) was too large. Try again with a smaller estimate. e. Repeat parts (c) and (d) successively for \(\epsilon=0.1,0.05,\) and \(0.001 .\) $$ f(x)=\frac{3 x^{2}-(7 x+1) \sqrt{x}+5}{x-1}, \quad x_{0}=1 $$

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