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Chapter 1: Problem 31

The value of the limit $\lim _{n \rightarrow \infty} \mathrm{n}^{2}(\sqrt[n]{a}-\sqrt[n+1]{a})(a>0)$ is (A) \(\ell\) n a (B) \(\mathrm{e}^{\mathrm{a}}\) (C) \(\mathrm{e}^{-\mathrm{a}}\) (D) none of these

Short Answer

Expert verified
(A) 0 (B) 1 (C) infinity (D) none of these Answer: (D) none of these

Step by step solution

01

Rewrite the expression as a single root

We have the expression, \(n^2(\sqrt[n]{a} - \sqrt[n+1]{a})\), which can be rewritten as \(n^2(\frac{a^{1/n} - a^{1/(n+1)}}{a^{1/(n+1)}})\) by taking the factor \(a^{1/(n+1)}\) out of the second term and dividing both terms by \(a^{1/(n+1)}\). Now, we can rewrite it as a single root expression: \(n^2(a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\)
02

Apply the chain rule for limits

Now, we can apply limit as \(n \to \infty\) on the given expression: \(\lim_{n\to\infty} n^2(a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\) With this expression, we can apply the chain rule for limits. We have two terms: \(n^2\) and \((a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\).
03

Find the limit for each term

For the first term: \(\lim_{n\to\infty} n^2 = \infty\) For the second term, we can simplify \(1/n - 1/(n+1)\): \(1/n - 1/(n+1) = \frac{1}{n(n+1)}\) Now, for the limit of the second term: \(\lim_{n\to\infty} (a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\) We can rewrite it as: \(\lim_{n\to\infty} (a^{1/(n+1)}(a^{\frac{1}{n(n+1)}} - 1))\) Now, we apply limit, and use the property that \(\lim_{x\to 0}(1+x)^{1/x} = e\). So, \(\lim_{n\to\infty} a^{\frac{1}{n(n+1)}} = e^{\ln a}\) And the limit of the second term is: \(\lim_{n\to\infty} (a^{1/(n+1)}(e^{\ln a} - 1)) = e^{\ln a}(e^{\ln a} - 1)\)
04

Multiply the limits

Now, we can multiply the limits of both terms: \(\lim_{n\to\infty} n^2(a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\) \(= (\lim_{n\to\infty} n^2)(\lim_{n\to\infty} a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\) \(= \infty (e^{\ln a}(e^{\ln a} - 1))\) Since \(a > 0\), \(e^{\ln a}(e^{\ln a} - 1)\) is finite. Therefore, the given expression is unbounded, and the limit does not exist. Thus, the correct answer is: (D) none of these

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Most popular questions from this chapter

$\lim _{n \rightarrow \infty}\left(\frac{\sqrt[n]{p}+\sqrt[n]{q}}{2}\right)^{n}\(, p, \)q>0$ is equal to (A) 1 (B) \(\sqrt{\mathrm{pq}}\) (C) pq (D) \(\frac{\mathrm{pq}}{2}\)

$\lim _{x \rightarrow 0} \frac{6 x^{2}(\cot x)(\operatorname{cosec} 2 x)}{\sec \left(\cos x+\pi \tan \left(\frac{\pi}{4 \sec x}\right)-1\right)}$ has the value equal to (A) 6 (B) \(-6\) (C) 0 (D) \(-3\)

If \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are real numbers then the value of $\lim _{t \rightarrow 0} \ln \left(\frac{1}{t} \int_{0}^{t}(1+a \sin b x)^{\mathrm{c} / x} d x\right)$ equals (A) \(a b c\) (B) \(\frac{a b}{c}\) (C) \(\frac{b c}{a}\) (D) \(\frac{\mathrm{ca}}{\mathrm{b}}\)

$\lim _{y \rightarrow 0}\left[\lim _{x \rightarrow \infty} \frac{\exp \left(\mathrm{x} \ell \mathrm{n}\left(1+\frac{\mathrm{ay}}{\mathrm{x}}\right)\right)-\exp \left(\mathrm{x} \& \mathrm{n}\left(1+\frac{\mathrm{by}}{\mathrm{x}}\right)\right)}{\mathrm{y}}\right]$ $\begin{array}{ll}\text { (A) } \mathrm{a}+\mathrm{b} & \text { (B) } \mathrm{a}-\mathrm{b}\end{array}$ (C) \(b-a \quad\) (D) \(-(a+b)\)

The value of $\lim _{x \rightarrow 0} \frac{(\tan (\\{x\\}-1)) \sin \\{x\\}}{\\{x\\}(\\{x\\}-1)}\(, where \)\\{x\\}$ denotes the fractional part function, is (A) is 1 (B) is tan 1 (C) is \(\sin 1\) (D) is non-existent
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