Question

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

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

Step-by-step solution

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: \(n^2(a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\)

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 \((a^{1/(n+1)}(a^{1/n - 1/(n+1)} - 1))\).

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: \(\lim_{n\to\infty} (a^{1/(n+1)}(e^{\ln a} - 1)) = e^{\ln a}(e^{\ln a} - 1)\)

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