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Chapter 2: Problem 26

Find the limits in Exercises 21鈥�36. $$ \lim _{t \rightarrow 0} \frac{2 t}{\tan t} $$

Short Answer

Expert verified
The limit is 2.

Step by step solution

01

Identify the Indeterminate Form

As \( t \rightarrow 0 \), both the numerator \(2t\) and the denominator \( \tan t \) approach 0. This creates an indeterminate form of type \( \frac{0}{0} \), indicating that L'H么pital's Rule might be suitable to solve the limit.
02

Apply L'H么pital's Rule

L'H么pital's Rule can be used when \( \lim_{t \rightarrow a} \frac{f(t)}{g(t)} = \frac{0}{0} \). The rule states that \( \lim_{t \rightarrow a} \frac{f(t)}{g(t)} = \lim_{t \rightarrow a} \frac{f'(t)}{g'(t)} \), assuming the limit on the right exists. For our function, differentiate the numerator and denominator: - Numerator: \( (2t)' = 2 \)- Denominator: \( (\tan t)' = \sec^2 t \)
03

Compute the New Limit

Using L'H么pital's Rule, we find\[\lim _{t \rightarrow 0} \frac{2 t}{\tan t} = \lim_{t \rightarrow 0} \frac{2}{\sec^2 t} = \lim_{t \rightarrow 0} 2 \cos^2 t\]
04

Evaluate the Limit

Since \( \cos 0 = 1 \), we have \[\lim_{t \rightarrow 0} 2 \cos^2 t = 2 \cdot 1^2 = 2.\]Thus, the limit is \(2\).

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Key Concepts

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

Understanding Limits
Limits are a fundamental concept in calculus used to understand the behavior of functions as they approach a certain point. In simpler terms, when we talk about the limit of a function as it approaches a particular value, we're interested in what value the function itself is getting closer to.

When you see a notation like \( \lim_{t \rightarrow 0} \), it translates to, "What value does the function get closer to as \( t \) gets nearer to 0?" This understanding allows us to analyze functions that are not necessarily defined at certain points.

  • Why Limits Matter: They help in understanding the continuity and behavior of functions.
  • How to Evaluate: Sometimes direct substitution works. If not, techniques like factoring, rationalizing, or L'H么pital's Rule are used.
For trigonometric functions, limits often prepare us for more complex calculus topics like derivatives and integrals.
Exploring Indeterminate Forms
Indeterminate forms are expressions that do not lead to a clear answer when you substitute a value directly into them. These forms typically occur in calculus when limits are being evaluated.

The expression \( \frac{0}{0} \) is one of the most common indeterminate forms. It is 'indeterminate' because it can potentially be any number, requiring additional work to find an actual limit.

  • Common Types: \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \).
  • Resolving Indeterminate Forms: L'H么pital's Rule is a popular tool. It allows us to differentiate the numerator and denominator separately to find a clearer limit.
In our exercise, \( \lim_{t \rightarrow 0} \frac{2t}{\tan t} \) equals \( \frac{0}{0} \), making L'H么pital's Rule appropriate to use.
Simplifying Trigonometric Limits with L'H么pital's Rule
Trigonometric limits often include functions like \( \sin t \), \( \cos t \), and \( \tan t \). These limits can be challenging when they result in indeterminate forms. To solve these limits, L'H么pital's Rule can be extremely useful.

L'H么pital's Rule states that if a limit results in an indeterminate form like \( \frac{0}{0} \), it can be re-evaluated using the derivatives of the numerator and denominator.

  • Application in Trigonometric Limits: Differentiating \( 2t \) gives \( 2 \) and differentiating \( \tan t \) gives \( \sec^2 t \).
  • Evaluating: Replace the original limit with \( \lim_{t \rightarrow 0} \frac{2}{\sec^2 t} = \lim_{t \rightarrow 0} 2 \cos^2 t \).
  • Conclusion: Since \( \cos 0 = 1 \), the limit evaluates to \( 2 \).
Through this method, trigonometric limits that might otherwise seem daunting become manageable and solvable.

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Most popular questions from this chapter

In Exercises \(5-10\) , find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$ y=\frac{1}{x^{3}}, \quad\left(-2,-\frac{1}{8}\right) $$

One-sided limits Let \(f(x)=\left\\{\begin{array}{ll}{x^{2} \sin (1 / x),} & {x<0} \\ {\sqrt{x},} & {x>0}\end{array}\right.\) Find \((\mathrm{a}) \lim _{x \rightarrow 0^{+}} f(x)\) and \((\mathrm{b}) \lim _{x \rightarrow 0}-f(x) ;\) then use limit definitions to verify your findings. (c) Based on your conclusions in parts (a) and \((b),\) can anything be said about \(\lim _{x \rightarrow 0} f(x) ?\) Give reasons for your answer.

Find \(\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+x}-\sqrt{x^{2}-x}\right)\)

Graph the rational functions in Exercises \(27-38 .\) Include the graphs and equations of the asymptotes and dominant terms. $$ y=\frac{x^{2}}{x-1} $$

Given \(\epsilon>0,\) find an interval \(I=(5,5+\delta), \delta>0,\) such that if \(x\) lies in \(I,\) then \(\sqrt{x-5}<\epsilon .\) What limit is being verified and what is its value?
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