Question

A a) \(a_{n}=\frac{\sqrt{n}}{n}\) b) \(a_{n}=\frac{5+n}{2 n}\) c) \(a_{n}=\left(-\frac{1}{4}\right)^{n}-1\) d) \(a_{n}=\frac{2}{n}+1\) e) \(a_{n}=\frac{n^{3}+1}{2 n^{3}+n^{2}+n}\) f) \(a_{n}=2+2^{-n}\) g) \(a_{n}=\frac{n^{2}-1}{(n+1)^{2}}+5\) B Berechnen Sie die folgenden Grenzwerte a) \(\lim _{x \rightarrow 0} \frac{x^{2}+1}{x-1}\) b) \(\lim _{x \rightarrow 2} \frac{1}{x}\) c) \(\lim _{x \rightarrow 0} \frac{x^{2}+10 x}{2 x}\) d) \(\lim _{x \rightarrow \infty} e^{-x}\)

Short answer

A(a): 0, A(b): 0.5, A(c): -1, A(d): 1, A(e): 0.5, A(f): 2, A(g): 6. B(a): -1, B(b): 0.5, B(c): 5, B(d): 0

Step-by-step solution

Solving part A

Solve each sequence's limit as n approaches infinity.

Solve part A(a)

Determine the limit of \( \frac{\sqrt{n}}{n} \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{n} = \lim_{n \rightarrow \infty} \frac{n^{1/2}}{n} = \lim_{n \rightarrow \infty} \frac{1}{n^{1/2}} = 0 \)

Solve part A(b)

Determine the limit of \( \frac{5+n}{2n} \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} \frac{5+n}{2n} = \lim_{n \rightarrow \infty} \left( \frac{5}{2n} + \frac{n}{2n} \right) = \lim_{n \rightarrow \infty} \left( \frac{5}{2n} + \frac{1}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2} \)

Solve part A(c)

Determine the limit of \( \left( -\frac{1}{4} \right)^n - 1 \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} \left( -\frac{1}{4} \right)^n - 1 = 0 - 1 = -1 \)

Solve part A(d)

Determine the limit of \( \frac{2}{n} + 1 \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} \frac{2}{n} + 1 = 0 + 1 = 1 \)

Solve part A(e)

Determine the limit of \( \frac{n^3 + 1}{2n^3 + n^2 + n} \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} \frac{n^3 + 1}{2n^3 + n^2 + n} = \lim_{n \rightarrow \infty} \frac{n^3(1+1/n^3)}{2n^3(1+1/n+1/n^2)} = \frac{1}{2} \)

Solve part A(f)

Determine the limit of \( 2 + 2^{-n} \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} 2 + 2^{-n} = 2 + 0 = 2 \)

Solve part A(g)

Determine the limit of \( \frac{n^2 - 1}{(n+1)^2} + 5 \) as \( n \rightarrow \infty \): \( \lim_{n \rightarrow \infty} \frac{n^2 - 1}{(n+1)^2} + 5 = \lim_{n \rightarrow \infty} \frac{n^2 (1 - \frac{1}{n^2})}{n^2 (1 + \frac{2}{n} + \frac{1}{n^2})} + 5 = \lim_{n \rightarrow \infty} \left( \frac{1 - \frac{1}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}} \right) + 5 = 1 + 5 = 6 \)

Solving part B

Solve each limit as x approaches the specified value.

Solve part B(a)

Determine the limit of \( \frac{x^2 + 1}{x - 1} \) as \( x \rightarrow 0 \): Since direct substitution results in a well-defined value, we get: \( \lim_{x \rightarrow 0} \frac{x^2 + 1}{x - 1} = \frac{0^2 + 1}{0 - 1} = \frac{1}{-1} = -1 \)

Solve part B(b)

Determine the limit of \( \frac{1}{x} \) as \( x \rightarrow 2 \): Since direct substitution results in a well-defined value, we get: \( \lim_{x \rightarrow 2} \frac{1}{x} = \frac{1}{2} = \frac{1}{2} \)

Solve part B(c)

Determine the limit of \( \frac{x^2 + 10x}{2x} \) as \( x \rightarrow 0 \): \( \lim_{x \rightarrow 0} \frac{x^2 + 10x}{2x} = \lim_{x \rightarrow 0} \frac{x(x+10)}{2x} = \lim_{x \rightarrow 0} \frac{x+10}{2} = \frac{0 + 10}{2} = 5 \)

Solve part B(d)

Determine the limit of \( e^{-x} \) as \( x \rightarrow \infty \): \( \lim_{x \rightarrow \infty} e^{-x} = 0 \)

Key concepts

Sequence Limits

Calculus often deals with sequences, which are ordered lists of numbers. Understanding sequence limits is crucial because it helps determine the behavior of sequences as they progress toward infinity. For example, consider the sequence given by \( a_{n} = \frac{\root{n}}{n} \). To find the limit as \(n \to \infty\), you must evaluate:
\[ \frac{\root{n}}{n} = \frac{n^{1/2}}{n} = \frac{1}{n^{1/2}} \to 0 \]
This shows the sequence approaches zero as \( n \) becomes very large. Similarly, in the other sequences provided, simplifying terms and using properties of limits help determine the asymptotic behavior.

Infinite Limits

Infinite limits explain what happens to a function or sequence as it approaches a certain value or becomes indefinitely large. In part A(f) of the exercise, this involves \( a_{n} = 2 + 2^{-n} \). As \( n \rightarrow \infty \), know that \( 2^{-n} \rightarrow 0 \), so the limit becomes:
\[ 2 + 0 = 2 \]
Infinite limits are common in sequences and functions where the main part of the expression becomes negligibly small or infinitely large, depending on the value of \(n\).

Calculus Step-By-Step Solutions

Step-by-step solutions are pivotal for understanding how to solve calculus problems correctly. It structures the solving approach, ensuring clarity and learning along the way. For instance, in part A(e) of the exercise where \( a_{n} = \frac{n^{3} + 1}{2n^{3} + n^{2} + n} \), breaking down the limit step-by-step helps show how each term behaves as \(n \rightarrow \infty\):
\[ \frac{n^{3}(1+\frac{1}{n^{3}})}{2n^{3}(1+\frac{1}{n}+\frac{1}{n^{2}})} = \frac{1+\frac{1}{n^{3}}}{2+\frac{1}{n^{2}}+\frac{1}{n}} \to \frac{1}{2} \]
This helps see why the result is \( \frac{1}{2} \). Following clear steps like this aids in understanding the meticulous process behind finding limits.

Limits at Infinity

Limits at infinity focus on what happens to a function as the variable grows very large or very small. Analyzing these limits helps in determining long-term behavior. For instance, consider part B(d) with \( \text{lim}_{x \rightarrow \text{infty}} e^{-x} \). Since the exponential function drops off rapidly to zero, the outcome is:
\[ e^{-x} \rightarrow 0 \]
Similarly, examining \( \text{lim}_{n \rightarrow \text{infty}} \frac{2}{n} + 1 \) in part A(d), as n becomes very large, we understand that:
\[ \text{lim}_{n \rightarrow \text{infty}} \frac{2}{n} + 1 = 0 + 1 = 1 \]
These conclusions help in realizing the behavior of functions or sequences in long-run scenarios.