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Chapter 3: Problem 10

Determine the following limits. (a) \(\lim \left((3 \sqrt{n})^{1 / 2 n}\right)\), (b) \(\lim \left((n+1)^{1 / \ln (n+1)}\right)\).

Short Answer

Expert verified
The limits are given as: (a) \( e^2 \) and (b) 2.

Step by step solution

01

Part (a): Rearrange terms

We want to rewrite \((3 \sqrt{n})^{1 / 2 n}\) into a form that uses the limit definition of e. This can be done by splitting apart the fraction within the exponent: \[ (3 \sqrt{n})^{1 / 2 n} = (3 \sqrt{n})^{\frac{1}{n} / 2} = (3^{1/n} \cdot n^{1/2n})^2 \].
02

Part (a): Evaluating the limit

Now we can evaluate: \[ \lim_{n\to\infty}(3^{1/n} \cdot n^{1/2n})^2 = \lim_{n\to\infty)(((1 + \frac{2}{n})^{n/2})^2\) As \(n\) approaches infinity, the expression \((1 + \frac{2}{n})^{n/2}\) approaches \(e^{2/2} = e\). Thus, the limit of the expression is \(e^2\).
03

Part (b): Rearrange terms

We do a similar rearrangement for \((n+1)^{1 / \ln (n+1)}\): \[ (n+1)^{1 / \ln (n+1)} = ((1 + \frac{1}{n})^{n})^{1 / \ln(1 + \frac{1}{n})} \]
04

Part (b): Evaluating the limit

Now we can evaluate: \[ \lim_{n\to 1} ((1 + \frac{1}{n})^{n})^{1 / \ln(1 + \frac{1}{n})} = e^{1/\ln 2} = 2 \] To get this result, we've used the fact that \(\ln 2 = \log e 2\), and the property that \(a^{\log a b} = b\), so \(e^{\log e 2} = 2\).

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Most popular questions from this chapter

Prove that \(\lim \left(x_{n}\right)=0\) if and only if \(\lim \left(\left|x_{n}\right|\right)=0\). Give an example to show that the convergence of \(\left(\left|x_{n}\right|\right)\) need not imply the convergence of \(\left(x_{n}\right)\).

If \(a>0, b>0\), show that lim \((\sqrt{(n+a)(n+b)}-n)=(a+b) / 2\).

Let \(x_{1}:=1\) and \(x_{n+1}:=\sqrt{2+x_{n}}\) for \(n \in \mathbb{N}\). Show that \(\left(x_{n}\right)\) converges and find the limit.

Show that (a) \(\lim \left(\frac{1}{\sqrt{n+7}}\right)=0\), (b) \(\lim \left(\frac{2 n}{n+2}\right)=2\), (c) \(\lim \left(\frac{\sqrt{n}}{n+1}\right)=0\), (d) \(\lim \left(\frac{(-1)^{n} n}{n^{2}+1}\right)=0\).

Let \(y_{1}:=\sqrt{p}\), where. \(p>0\), and \(y_{n+1}:=\sqrt{p+y_{n}}\) for \(n \in \mathbb{N}\). Show that \(\left(y_{n}\right)\) converges and find the limit. [Hint: One upper bound is \(1+2 \sqrt{p} .]\)
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