/*! 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 12 \(5-64\) Find the limit. Use I'H... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(5-64\) Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{t \rightarrow 0} \frac{e^{3 t}-1}{t}$$

Short Answer

Expert verified
The limit is 3.

Step by step solution

01

Check the Indeterminate Form

First, substitute the value of \( t = 0 \) into the expression \( \frac{e^{3t} - 1}{t} \) to see if it forms an indeterminate expression like \( \frac{0}{0} \). This will allow us to determine if L'Hôpital's Rule can be applied.\[\frac{e^{3(0)} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}\]The limit gives an indeterminate form \( \frac{0}{0} \), so we can proceed with L'Hôpital's Rule.
02

Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit \( \frac{f(t)}{g(t)} \) as \( t \) approaches a point results in an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then the limit can be found by taking the derivatives of the numerator and the denominator.Find the derivatives of the numerator and the denominator:\( f(t) = e^{3t} - 1 \), its derivative is \( f'(t) = 3e^{3t} \)\( g(t) = t \), its derivative is \( g'(t) = 1 \)Substitute these into L'Hôpital's Rule:\[\lim_{t \rightarrow 0} \frac{e^{3t} - 1}{t} = \lim_{t \rightarrow 0} \frac{3e^{3t}}{1}\]
03

Evaluate the Limit After Differentiation

Now, substitute \( t = 0 \) into the differentiated expression:\[\lim_{t \rightarrow 0} 3e^{3t} = 3e^{3(0)} = 3 \times 1 = 3\]The limit of \( \frac{e^{3t} - 1}{t} \) as \( t \rightarrow 0 \) is therefore 3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Limits
In calculus, limits are fundamental to understanding how values behave as they approach a certain point. The concept of a limit is used to determine the value that a function approaches as the input approaches a specified value. When evaluating limits, you're essentially asking what the function's value gets infinitesimally close to, even if it doesn't actually reach that value.
To compute a limit, you often substitute the value into the function. However, if direct substitution leads to indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you must use other methods, like L'Hôpital's Rule or algebraic manipulation, to find the limit.

Limits are crucial in calculus because they form the basis of derivatives and integrals. They help in understanding behaviors of functions at points where they might not be clearly defined by just looking at the formula.
L'Hôpital's Rule
L'Hôpital's Rule is a valuable tool in calculus for resolving indeterminate forms. Named after the French mathematician Guillaume de l'Hôpital, this rule provides a way to evaluate limits of indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) by using derivatives.
L'Hôpital's Rule states that if you have a limit of the form \( \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \) and this results in an indeterminate form, you can differentiate the numerator and the denominator separately and re-evaluate the limit:
  • The limit of \( \frac{f(x)}{g(x)} \) becomes \( \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \) given that \( f'(x) \) and \( g'(x) \) are continuous.
  • Applying this rule is straightforward but it requires that the original forms must first result in indeterminacies like \( \frac{0}{0} \).
By understanding and applying L'Hôpital's Rule, you can evaluate complex limits that would otherwise be difficult to handle with basic substitution or algebraic manipulation.
Indeterminate Forms
Indeterminate forms occur when evaluating a limit results in an ambiguous expression, such as \( \frac{0}{0} \), \( \infty - \infty \), or \( 0 \times \infty \). These forms don't provide enough information to determine a specific limit value directly, making further analysis necessary.

When facing an indeterminate form, you might use algebraic techniques or calculus tools such as L'Hôpital's Rule to find the limit. Specifically:
  • In the case of \( \frac{0}{0} \), L'Hôpital's Rule can help by allowing you to take derivatives of the numerator and denominator.
  • For other forms, it might be necessary to rearrange or simplify the expression, apply trigonometric identities, or factor terms to find a viable limit.
Recognizing when an expression is an indeterminate form is key. It signals the need for more advanced techniques to accurately determine the behavior of a function as it approaches a particular point.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Use Newton's method with \(\mathrm{x}_{1}=1\) to find the root of the equation \(\mathrm{x}^{3}-\mathrm{x}=1\) correct to six decimal places. b) Solve the equation in part (a) using \(x_{1}=0.6\) as the initial approximation. (c) Solve the equation in part (a) using \(x_{1}=0.57 .\) (You defi- nitely need a programmable calculator for this part.) (d) Graph \(f(x)=x^{3}-x-1\) and its tangent lines at \(x_{1}=1\) \(0.6,\) and 0.57 to explain why Newton's method is so sen- sitive to the value of the initial approximation.v

Produce graphs of \(f\) that reveal all the important aspects of the curve. In particular, you should use graphs of \(f^{\prime}\) and \(f "\) to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. \(f(x)=x^{2}-4 x+7 \cos x, \quad-4 \leqslant x \leqslant 4\)

since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. A rain-drop has an initial downward velocity of 10 \(\mathrm{m} / \mathrm{s}\) and its downward acceleration is $$a=\left\\{\begin{array}{ll}{9-0.9 t} & {\text { if } 0 \leq t \leq 10} \\\ {0} & {\text { if } t>10}\end{array}\right.$$ If the raindrop is initially 500 \(\mathrm{m}\) above the ground, how long does it take to fall?

Describe how the graph of \(f\) varies as \(c\) varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when c changes. You should also identify any transitional values of \(c\) at which the basic shape of the curve changes. \(f(x)=\ln \left(x^{2}+c\right)\)

Produce graphs of \(f\) that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to find these intervals exactly. \(f(x)=\frac{1}{x^{8}}-\frac{2 \times 10^{8}}{x^{4}}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.