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\( \lim _{n \rightarrow \infty}\left(2^{n}+3^{n}+11^{n}\right)^{\frac{1}{n}} \) is equal to |
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Answer» \(L = \lim\limits_{n \to \infty} (2^n + 3^n + 11^n) \) \((\infty^0 - case)\) \(log\, L = \lim\limits_{n \to \infty} \frac1n (log (2^n + 3^n + 11^n) )\) \(\left(\frac\infty\infty - case\right)\) \(= \lim\limits_{n \to \infty } \frac{2^n log2 + 3^n log2 + 11^n log11}{2^n + 3^n + 11^n}\) (By using D.L.H. Rule) \(=\lim\limits_{n\to\infty} \cfrac{11^n\left(\left(\frac2{11}\right)^n log2 +\left(\frac3{11}\right)^n log2 + log 11 \right)}{11^n \left(\left(\frac2{11}\right)^n + \left(\frac3{11}\right)^n + 1\right)}\) \(=log11\left(\because\lim\limits_{n\to\infty}\left(\frac2{11}\right)^n =\lim\limits_{n\to\infty}\left(\frac3{11}\right)^n = 0 \right)\) \(\therefore L = 11\) ⇒ \(\lim\limits_{n\to \infty}(2^n + 3^n + 11^n)^\frac1n = 11\) |
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