Chapter 9: Problem 698
\(\lim _{\mathrm{x} \rightarrow 1}\left[\left(\mathrm{x}^{365}-365 \mathrm{x}+364\right) /(\mathrm{x}-1)^{2}\right]=?\) (a) 66,430 (b) 64,340 (c) 66,630 (d) 64,430
Short Answer
Expert verified
The short answer is: \(\lim_{x \rightarrow 1} \frac{x^{365} - 365x + 364}{(x-1)^{2}} = 66,430\).
Step by step solution
01
Identify Indeterminate Form
As x approaches 1, we have:
\[
\lim_{x \rightarrow 1} \frac{x^{365} - 365x + 364}{(x-1)^{2}}
\]
Plugging in x = 1 to the numerator and the denominator, we get:
\[
\frac{1^{365} - 365 \cdot 1 + 364}{(1-1)^{2}} = \frac{0}{0}
\]
Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hopital's Rule.
02
Apply L'Hopital's Rule
Differentiate both the numerator and the denominator with respect to x:
\[
\frac{d}{dx}(x^{365} - 365x + 364) = 365x^{364} - 365
\]
\[
\frac{d}{dx}((x-1)^2) = 2(x-1)^1
\]
Now apply L'Hopital's Rule:
\[
\lim_{x \rightarrow 1} \frac{365x^{364} - 365}{2(x-1)}
\]
03
Evaluate the Limit Again
Plug in x = 1 to the new expression:
\[
\frac{365 \cdot 1^{364} - 365}{2(1-1)} = \frac{365 - 365}{0}
\]
The limit is still in indeterminate form. Apply L'Hopital's Rule again.
04
Apply L'Hopital's Rule a Second Time
Differentiate both the numerator and the denominator with respect to x again:
\[
\frac{d}{dx}(365x^{364} - 365) = 365 \cdot 364 \cdot x^{363}
\]
\[
\frac{d}{dx}(2(x-1)) = 2
\]
Now apply L'Hopital's Rule again:
\[
\lim_{x \rightarrow 1} \frac{365 \cdot 364 \cdot x^{363}}{2}
\]
05
Evaluate the Limit for the Final Time
Plug in x = 1 to the resulting expression:
\[
\frac{365 \cdot 364 \cdot 1^{363}}{2} = \frac{365 \cdot 364}{2} = 365 \cdot 182 = 66430
\]
So the limit is:
\[
\lim_{x \rightarrow 1} \frac{x^{365} - 365x + 364}{(x-1)^{2}} = 66430
\]
The correct answer is (a) 66,430.
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 Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
When dealing with calculus, specifically limits, we often encounter expressions that lead to indeterminate forms. An indeterminate form is an undefined expression that does not have a unique limit or may have multiple limits. The most common indeterminate form is \( \frac{0}{0} \), as seen in the provided exercise. Others include \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), \( 1^\infty \), and \( 0^0 \).
Understanding indeterminate forms is crucial since they inform us that more investigation is needed to find the limit of a function. It indicates that direct substitution will not work, and we must apply techniques like factoring, rationalization, or, as in the given problem, L'Hopital's Rule, to evaluate these limits properly.
Understanding indeterminate forms is crucial since they inform us that more investigation is needed to find the limit of a function. It indicates that direct substitution will not work, and we must apply techniques like factoring, rationalization, or, as in the given problem, L'Hopital's Rule, to evaluate these limits properly.
Limits
A limit in calculus is essentially the value that a function approaches as the input approaches some point. Limits are foundational to the study of calculus because they underpin the definitions of continuity, derivatives, and integrals. Finding the limit of a function like \( \lim_{{x \rightarrow a}} f(x) \) often involves seeing what value f(x) will approach as x gets arbitrarily close to a.
However, when the function produces an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), direct substitution isn't possible. That's where L'Hopital's Rule comes into play—it gives us a reliable method to evaluate many limits that at first seem uncertain.
However, when the function produces an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), direct substitution isn't possible. That's where L'Hopital's Rule comes into play—it gives us a reliable method to evaluate many limits that at first seem uncertain.
Calculus for JEE
Calculus is a significant part of the Mathematics syllabus for the Joint Entrance Examination (JEE), which is an engineering entrance assessment conducted in India. It's imperative for students preparing for the JEE to grasp concepts like differentiation and limits comprehensively, as they are the bedrock of advanced studies in physics and engineering.
L'Hopital's Rule, a topic often featured in JEE prep materials, is especially useful for addressing limit problems that result in indeterminate forms. Mastering this rule not only aids in solving textbook problems but also prepares students for the conceptual understanding required in higher education and practical applications in engineering.
L'Hopital's Rule, a topic often featured in JEE prep materials, is especially useful for addressing limit problems that result in indeterminate forms. Mastering this rule not only aids in solving textbook problems but also prepares students for the conceptual understanding required in higher education and practical applications in engineering.
Differentiation
In calculus, differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes with respect to a change in its input value. Differentiation is a powerful tool used to solve a myriad of problems, ranging from simple motion to optimizing complex systems.
Within the context of L'Hopital's Rule, differentiation is used to simplify the evaluation of limits involving indeterminate forms. By taking the derivative of both the numerator and the denominator separately, we can convert an indeterminate form to a determinate one, which can then be more easily evaluated by direct substitution, as illustrated in the exercise's step-by-step solution.
Within the context of L'Hopital's Rule, differentiation is used to simplify the evaluation of limits involving indeterminate forms. By taking the derivative of both the numerator and the denominator separately, we can convert an indeterminate form to a determinate one, which can then be more easily evaluated by direct substitution, as illustrated in the exercise's step-by-step solution.