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How would I go about finding the following limit:

$$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}}$$

PTDS
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david david
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3 Answers3

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$$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}} = \lim_{n \to \infty} (2^n(1-n^2 2^{-n}))^{\frac{1}{n}} = 2 \lim_{n \to \infty} (1-n^2 2^{-n}))^{\frac{1}{n}} = 2$$

PTDS
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After taking logarithm with base $e$, it suffices to calculate: $$\lim_{x\rightarrow \infty}\frac{\mathrm{ln}(2^x-x^2)}{x}=\lim_{x\rightarrow\infty}\frac{\mathrm{ln}2 \cdot 2^{x} - 2x}{2^x - x^2}= \lim_{x\rightarrow\infty}\frac{\mathrm{ln}2 - \frac{2x}{2^{x}}}{1 - \frac{x^2}{2^x}}=\mathrm{ln}2.$$

Hence the original limit is 2.

Note. We used L'Hôpital's rule in the first equality.

Rigid AOE2
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$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}}=\lim_{n \to \infty} (2^n)^{\frac{1}{n}}(1-\frac{1}{\frac{2^n}{n^2}})^{(\frac{1}{n})(\frac{-2^n}{n^2})(\frac{-n^2}{2^n})}=\lim_{n \to \infty}2(e)^{\frac{-n}{2^n}}=2$.
The idea is to regain the special limit=$ \lim_{n \to \infty} (1+\frac{1}{n})^n=e$

gui
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