/*! 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 80 Evaluate the given limit. $$ \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the given limit. $$ \lim _{n \rightarrow \infty}\left(1+\frac{1}{n^{2}}\right)^{n^{2}} $$

Short Answer

Expert verified
The limit is \( e \).

Step by step solution

01

Understanding the Limit

We are asked to evaluate \( \lim _{n \rightarrow \infty}\left(1+\frac{1}{n^{2}}\right)^{n^{2}} \). This expression resembles the standard form of the exponential limit, specifically \( \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e \). However, instead of \(n\), we have \(n^2\). We can use this similarity to our advantage.
02

Apply the Exponential Limit Transformation

Rewrite \( \left(1+\frac{1}{n^{2}}\right)^{n^{2}} \) in a form that resembles the exponential limit: \( \left(1+\frac{1}{n^{2}}\right)^{n^{2}} = \left(\left(1+\frac{1}{n^{2}}\right)^{n^{2}}\right)^{1} = \left(1+\frac{1}{n^{2}}\right)^{n^{2}} \). Thus, the expression can be seen as \( e^{\lim_{n \to \infty} n^2 \ln\left(1 + \frac{1}{n^2}\right)} \).
03

Evaluate \( \lim_{n \to \infty} n^2 \ln\left(1 + \frac{1}{n^2}\right) \)

We need to find \( \lim_{n \to \infty} n^2 \ln\left(1 + \frac{1}{n^2}\right) \). Using the approximation \( \ln(1 + x) \approx x \) when \( x \to 0 \), we have \( \ln\left(1 + \frac{1}{n^2}\right) \approx \frac{1}{n^2} \) for large \( n \).
04

Simplify Limit Expression

Substitute the approximation into the limit: \( n^2 \ln\left(1 + \frac{1}{n^2}\right) \approx n^2 \cdot \frac{1}{n^2} = 1 \). Thus, \( \lim_{n \to \infty} n^2 \ln\left(1 + \frac{1}{n^2}\right) = 1 \).
05

Final Limit Evaluation

With the derived result, we find that: \( \lim_{n \to \infty}\left(1+\frac{1}{n^{2}}\right)^{n^{2}} = e^1 = e \). Thus, the original limit evaluates to \( e \).

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.

Exponential Limit
The concept of an exponential limit often appears in calculus, especially when dealing with sequences and series that tend toward infinity or involving powers of expressions. Limits of the form \( \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \) are particularly significant. This specific form converges to the mathematical constant \( e \), approximately equal to 2.71828.

Here's why this is important:
  • It describes the situation where a quantity approaches a certain value infinitely many times but never quite achieves it.
  • The expression \( \left(1 + \frac{1}{x}\right)^x \) models growth that is compounded continuously such as in finance and natural growth processes.
When calculating limits involving exponential forms, identifying similarities between the expression you have and the standard exponential limit form is key.

In the exercise, by recognizing \( \left(1+\frac{1}{n^{2}}\right)^{n^{2}} \) as an exponential limit transformation, we can simplify complicated expressions by turning them into simpler components directly related to the constant \( e \). This approach allows us to harness a well-known result to compute limits with ease.
Approximation Techniques
Often in calculus, especially when dealing with limits, exact values are not directly accessible. Here, approximation techniques become indispensable tools. One of the most common techniques used is the approximation of functions at infinitesimally small or large values.

In this exercise, the function \( \ln(1 + x) \) is approximated by \( x \) when \( x \to 0 \). This approximation \( \ln(1 + x) \approx x \) saves us time and simplifies the process of finding limits.

Why does this work?
  • The Taylor series expansion for \( \ln(1 + x) \) around 0 starts as \( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \) , but for very small \( x \), higher powers (like \( x^2 \) and \( x^3 \)) become negligible. Hence, \( \ln(1 + x) \approx x \).
  • This approximation is valid because the error becomes vanishingly small, making \( x \) the most significant term for a short interval.
When applying this to our initial problem, we simplified \( \ln\left(1 + \frac{1}{n^2} \right) \approx \frac{1}{n^2} \). With this, the computation becomes straightforward as the complexity of handling a logarithmic expression dissolves.
Infinity in Calculus
Infinity is a fundamental concept in calculus that refers to a process that continues indefinitely without ever reaching an endpoint. It's represented by the symbol \( \infty \) and frequently appears when calculating limits, derivatives, and integrals.

In the context of our specific problem, as \( n \to \infty \), we explore how some expressions behave or "settle down" as they grow larger.
  • Infinity allows us to explore the behavior of functions as they increase or decrease without bound.
  • Understanding this behavior is crucial when making predictions or analyzing long-term trends, such as determining whether a sequence converges to a finite limit or diverges.
For instance, examining \( \left(1+\frac{1}{n^{2}}\right)^{n^{2}} \), as \( n \) approaches infinity, we are essentially observing how the structure of an expression stabilizes to form a specific outcome.

The concept of infinity thus not only helps in theoretical aspects of calculus but also equips us with the analytical tools necessary to draw conclusions from complex, dynamic systems.

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

Radiocarbon Dating Two isotopes of carbon, \({ }^{12} \mathrm{C},\) which is stable, and \({ }^{14} \mathrm{C},\) which decays exponentially with a 5700 -year half-life, are found in a known fixed ratio in living matter. After death, carbon is no longer metabolized, and the amount \(m(t)\) of \({ }^{14} \mathrm{C}\) decreases due to radioactive decay. In the analysis of a sample performed \(T\) years after death, the mass of \({ }^{12} \mathrm{C}\), unchanged since death, can be used to determine the mass \(m_{0}\) of \({ }^{14} \mathrm{C}\) that the sample had at the moment of death. The time \(T\) since death can then be calculated from the law of exponential decay and the measurement of \(m(T)\). Use this information for solving Exercises \(95-98\) In \(1994,\) a parka-clad mummified body of a girl was found in a subterranean meat cellar near Barrow, Alaska. Radiocarbon analysis showed that the girl died around ce 1200 . What percentage of \(m_{0}\) was the amount of \({ }^{14} \mathrm{C}\) in the mummy?

In Exercises \(47-56,\) use your intuition to decide whether the limit exists. Justify your answer by using the rigorous definition of limit. $$ \begin{array}{l} \lim _{x \rightarrow 3} f(x) \text { where } \\ \qquad f(x)=\left\\{\begin{array}{cl} x^{2} & \text { if } x \leq 0 \\ 1-x^{2} & \text { if } x>0 \end{array}\right. \end{array} $$

In Exercises \(47-56,\) use your intuition to decide whether the limit exists. Justify your answer by using the rigorous definition of limit. $$ \lim _{x \rightarrow-7}(|x|-3 x) $$

A particle moves on an axis. Its position \(p(t)\) at time \(t\) is given. For a positive \(h,\) the average velocity over the time interval \([2,2+h]\) is \(\bar{v}(h)=\frac{p(2+h)-p(2)}{h}\) a. Numerically determine \(v_{0}=\lim _{h \rightarrow 0+} \bar{v}(h)\). b. How small does \(h\) need to be for \(\bar{v}(h)\) to be between \(v_{0}\) and \(v_{0}+0.1 ?\) c. How small does \(h\) need to be for \(\bar{v}(h)\) to be between \(v_{0}\) and \(v_{0}+0.01 ?\) $$ p(t)=(2 t)^{3 / 2}-2 t $$

Plot \(y=\exp (x)\) for \(0 \leq x \leq 2\). Let \(P(c)\) denote the point \((c, \exp (c))\) on the graph. The purpose of this exercise is to graphically explore the relationship between \(\exp (c)\) and the slope of the tangent line at \(P(c) .\) For \(c=1 / 2,1\), and \(3 / 2,\) calculate the slope \(m(c)\) of the secant line that passes through the pair of points \(P(c-0.001)\) and \(P(c+0.001) .\) For each \(c,\) calculate \(|\exp (c)-m(c)|\) to see that \(m(c)\) is a good approximation of \(\exp (c) .\) Add the three secant lines to your viewing window. For each of \(c=1 / 2,1,\) and \(3 / 2,\) add to the viewing window the line through \(P(c)\) with slope \(\exp (c)\). As we will see in Chapter \(3,\) these are the tangent lines at \(P(1 / 2), P(1)\) and \(P(3 / 2)\). It is likely that they cannot be distinguished from the secant lines in your plot.

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.