Limits Of Sequences A Detailed Mathematical Analysis

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Delving into the realm of limits, this problem focuses on evaluating the limit of a rational function as n approaches infinity. Understanding how to tackle such limits is crucial in calculus and real analysis. To effectively solve this, we need to analyze the dominant terms in both the numerator and the denominator. In this particular expression, we have a polynomial in the numerator and a polynomial in the denominator. The key strategy here is to divide both the numerator and the denominator by the highest power of n present in the denominator, which in this case is n4. This step allows us to simplify the expression and identify the terms that will approach zero as n grows infinitely large.

By dividing each term by n4, we transform the expression into a more manageable form:

lim⁡n→∞1000n3n4+3n2n40.001n4n4−100n3n4+1n4{\lim_{n \to \infty} \frac{\frac{1000n^3}{n^4} + \frac{3n^2}{n^4}}{\frac{0.001n^4}{n^4} - \frac{100n^3}{n^4} + \frac{1}{n^4}}}

Simplifying further, we get:

lim⁡n→∞1000n+3n20.001−100n+1n4{\lim_{n \to \infty} \frac{\frac{1000}{n} + \frac{3}{n^2}}{0.001 - \frac{100}{n} + \frac{1}{n^4}}}

As n approaches infinity, terms like 1000/n, 3/n2, 100/n, and 1/n4 all approach zero. This is a fundamental concept in limit evaluation: any constant divided by a term that goes to infinity will itself go to zero. Applying this understanding, the expression simplifies significantly.

We are left with:

0+00.001−0+0=00.001=0{\frac{0 + 0}{0.001 - 0 + 0} = \frac{0}{0.001} = 0}

Therefore, the limit of the given expression as n approaches infinity is 0. This result indicates that the denominator grows much faster than the numerator, causing the overall fraction to diminish towards zero. This method of dividing by the highest power of n is a cornerstone technique in evaluating limits of rational functions at infinity.

In summary, to solve this limit problem, the following key steps are undertaken. First, the identification of the highest power of n in the denominator is crucial. Second, dividing both the numerator and denominator by this highest power simplifies the expression. Finally, recognizing and applying the principle that constants divided by increasingly large numbers approach zero allows for the straightforward evaluation of the limit. This problem underscores the importance of algebraic manipulation and a solid understanding of limit behavior in calculus.

This problem presents another intriguing limit involving polynomial expressions. This time, we are dealing with the difference and sum of binomial expressions raised to the fourth power. To evaluate this limit as n approaches infinity, we need to first expand these binomials and simplify the expression. The expansion of (n+1)4 and (n-1)4 will involve using the binomial theorem or Pascal's triangle, which provides a systematic way to determine the coefficients in the expansion. Once expanded, we can identify and combine like terms to simplify both the numerator and the denominator.

Expanding (n+1)4, we get:

(n+1)4=n4+4n3+6n2+4n+1{(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n + 1}

Similarly, expanding (n-1)4, we get:

(n−1)4=n4−4n3+6n2−4n+1{(n-1)^4 = n^4 - 4n^3 + 6n^2 - 4n + 1}

Now, substituting these expansions into the original expression, we have:

lim⁡n→∞(n4+4n3+6n2+4n+1)−(n4−4n3+6n2−4n+1)(n4+4n3+6n2+4n+1)+(n4−4n3+6n2−4n+1){\lim_{n \to \infty} \frac{(n^4 + 4n^3 + 6n^2 + 4n + 1) - (n^4 - 4n^3 + 6n^2 - 4n + 1)}{(n^4 + 4n^3 + 6n^2 + 4n + 1) + (n^4 - 4n^3 + 6n^2 - 4n + 1)}}

Simplifying the numerator by subtracting the second polynomial from the first, we get:

n4+4n3+6n2+4n+1−n4+4n3−6n2+4n−1=8n3+8n{n^4 + 4n^3 + 6n^2 + 4n + 1 - n^4 + 4n^3 - 6n^2 + 4n - 1 = 8n^3 + 8n}

Simplifying the denominator by adding the two polynomials, we get:

n4+4n3+6n2+4n+1+n4−4n3+6n2−4n+1=2n4+12n2+2{n^4 + 4n^3 + 6n^2 + 4n + 1 + n^4 - 4n^3 + 6n^2 - 4n + 1 = 2n^4 + 12n^2 + 2}

So, the limit becomes:

lim⁡n→∞8n3+8n2n4+12n2+2{\lim_{n \to \infty} \frac{8n^3 + 8n}{2n^4 + 12n^2 + 2}}

Similar to the previous problem, we divide both the numerator and the denominator by the highest power of n in the denominator, which is n4:

lim⁡n→∞8n3n4+8nn42n4n4+12n2n4+2n4{\lim_{n \to \infty} \frac{\frac{8n^3}{n^4} + \frac{8n}{n^4}}{\frac{2n^4}{n^4} + \frac{12n^2}{n^4} + \frac{2}{n^4}}}

Simplifying, we get:

lim⁡n→∞8n+8n32+12n2+2n4{\lim_{n \to \infty} \frac{\frac{8}{n} + \frac{8}{n^3}}{2 + \frac{12}{n^2} + \frac{2}{n^4}}}

As n approaches infinity, the terms 8/n, 8/n3, 12/n2, and 2/n4 all approach zero. Thus, the limit becomes:

0+02+0+0=02=0{\frac{0 + 0}{2 + 0 + 0} = \frac{0}{2} = 0}

Therefore, the limit of the given expression as n approaches infinity is 0. This result again indicates that the denominator grows faster than the numerator, causing the fraction to approach zero.

The key takeaway from this problem is the importance of expanding binomial expressions and simplifying the resulting polynomials before evaluating the limit. Dividing by the highest power of n and understanding the behavior of terms as n goes to infinity are crucial steps in solving such problems. This approach is fundamental in the study of limits and series in calculus.

In this limit problem, we continue our exploration of limits involving polynomial expressions. Similar to the previous question, we need to expand the binomial terms raised to the fourth power. However, this time, the expressions inside the binomials are slightly more complex, involving terms like (2n + 1) and (n - 1). The same principle applies: we must expand, simplify, and then analyze the dominant terms as n approaches infinity. This type of problem reinforces the importance of algebraic manipulation and attention to detail in calculus.

First, let's expand (2n + 1)4. Using the binomial theorem or Pascal's triangle, we get:

(2n+1)4=(2n)4+4(2n)3(1)+6(2n)2(1)2+4(2n)(1)3+14{(2n+1)^4 = (2n)^4 + 4(2n)^3(1) + 6(2n)^2(1)^2 + 4(2n)(1)^3 + 1^4}

Simplifying this, we have:

(2n+1)4=16n4+32n3+24n2+8n+1{(2n+1)^4 = 16n^4 + 32n^3 + 24n^2 + 8n + 1}

We already know the expansion of (n - 1)4 from the previous problem:

(n−1)4=n4−4n3+6n2−4n+1{(n-1)^4 = n^4 - 4n^3 + 6n^2 - 4n + 1}

Now, we substitute these expansions into the original limit expression:

lim⁡n→∞(16n4+32n3+24n2+8n+1)−(n4−4n3+6n2−4n+1)(16n4+32n3+24n2+8n+1)+(n4−4n3+6n2−4n+1){\lim_{n \to \infty} \frac{(16n^4 + 32n^3 + 24n^2 + 8n + 1) - (n^4 - 4n^3 + 6n^2 - 4n + 1)}{(16n^4 + 32n^3 + 24n^2 + 8n + 1) + (n^4 - 4n^3 + 6n^2 - 4n + 1)}}

Simplifying the numerator by subtracting the second polynomial from the first, we get:

16n4+32n3+24n2+8n+1−n4+4n3−6n2+4n−1=15n4+36n3+18n2+12n{16n^4 + 32n^3 + 24n^2 + 8n + 1 - n^4 + 4n^3 - 6n^2 + 4n - 1 = 15n^4 + 36n^3 + 18n^2 + 12n}

Simplifying the denominator by adding the two polynomials, we get:

16n4+32n3+24n2+8n+1+n4−4n3+6n2−4n+1=17n4+28n3+30n2+4n+2{16n^4 + 32n^3 + 24n^2 + 8n + 1 + n^4 - 4n^3 + 6n^2 - 4n + 1 = 17n^4 + 28n^3 + 30n^2 + 4n + 2}

So, the limit becomes:

lim⁡n→∞15n4+36n3+18n2+12n17n4+28n3+30n2+4n+2{\lim_{n \to \infty} \frac{15n^4 + 36n^3 + 18n^2 + 12n}{17n^4 + 28n^3 + 30n^2 + 4n + 2}}

Again, we divide both the numerator and the denominator by the highest power of n in the denominator, which is n4:

lim⁡n→∞15n4n4+36n3n4+18n2n4+12nn417n4n4+28n3n4+30n2n4+4nn4+2n4{\lim_{n \to \infty} \frac{\frac{15n^4}{n^4} + \frac{36n^3}{n^4} + \frac{18n^2}{n^4} + \frac{12n}{n^4}}{\frac{17n^4}{n^4} + \frac{28n^3}{n^4} + \frac{30n^2}{n^4} + \frac{4n}{n^4} + \frac{2}{n^4}}}

Simplifying, we get:

lim⁡n→∞15+36n+18n2+12n317+28n+30n2+4n3+2n4{\lim_{n \to \infty} \frac{15 + \frac{36}{n} + \frac{18}{n^2} + \frac{12}{n^3}}{17 + \frac{28}{n} + \frac{30}{n^2} + \frac{4}{n^3} + \frac{2}{n^4}}}

As n approaches infinity, the terms 36/n, 18/n2, 12/n3, 28/n, 30/n2, 4/n3, and 2/n4 all approach zero. Thus, the limit becomes:

15+0+0+017+0+0+0+0=1517{\frac{15 + 0 + 0 + 0}{17 + 0 + 0 + 0 + 0} = \frac{15}{17}}

Therefore, the limit of the given expression as n approaches infinity is 15/17. This result shows that as n grows infinitely large, the ratio of the two polynomials approaches a constant value, indicating that they grow at a similar rate.

This problem highlights the importance of careful expansion and simplification of polynomial expressions. Dividing by the highest power of n remains a key technique, allowing us to focus on the constant terms that determine the limit. The problem reinforces the understanding of how polynomial growth rates affect limits as n approaches infinity, a critical concept in calculus and analysis.

This entry appears to be an incomplete limit problem, as the expression for which the limit needs to be evaluated is missing. To provide a comprehensive discussion, we would need the complete expression. However, we can still use this as an opportunity to discuss general strategies and considerations when approaching limit problems, especially those involving sequences and functions as n approaches infinity.

When dealing with limits as n approaches infinity, several common techniques and concepts come into play. As demonstrated in the previous problems, one of the most crucial techniques is to identify the dominant terms in the expression. For rational functions (ratios of polynomials), this often involves dividing both the numerator and the denominator by the highest power of n present in the expression. This simplification allows us to focus on the terms that will significantly impact the limit's value as n becomes very large.

Another critical concept is understanding the behavior of basic functions as n approaches infinity. For example, powers of n (e.g., n2, n3) will grow without bound, while terms of the form 1/nk (where k is a positive integer) will approach zero. Exponential functions (e.g., 2n) grow much faster than polynomial functions, and logarithmic functions (e.g., log n) grow much slower.

In cases involving more complex expressions, such as those with radicals, trigonometric functions, or combinations thereof, additional techniques may be necessary. These might include:

  1. Rationalization: If the expression involves square roots or other radicals, rationalizing the numerator or denominator can help simplify the expression and reveal the limit.
  2. Trigonometric Identities: If the expression involves trigonometric functions, using trigonometric identities can help rewrite the expression in a more manageable form.
  3. L'Hôpital's Rule: If the limit results in an indeterminate form (such as 0/0 or ∞/∞), L'Hôpital's Rule can be applied. This rule states that the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives, provided the limit of the derivatives exists.
  4. Squeeze Theorem (or Sandwich Theorem): If the function is bounded between two other functions whose limits are known, the Squeeze Theorem can be used to determine the limit of the original function.

Without the complete expression for the current problem, it's challenging to provide a specific solution. However, the general principles outlined here provide a framework for approaching a wide range of limit problems. Remember to carefully analyze the expression, identify the dominant terms, and apply appropriate techniques to simplify and evaluate the limit.

In conclusion, while problem 252 is incomplete, the discussion underscores the importance of a systematic approach to limit problems. Identifying the appropriate techniques and understanding the behavior of different functions as n approaches infinity are crucial skills in calculus and mathematical analysis. This comprehensive approach ensures that even complex limits can be tackled effectively.

Conclusion

This exploration of limits demonstrates fundamental techniques in calculus. From dividing by the highest power of n to expanding binomial expressions, these methods are crucial for understanding the behavior of functions as they approach infinity. Understanding these concepts provides a strong foundation for further studies in calculus and analysis.