How to evaluate the limit as it approaches infinity
$$\lim_{x \to \infty} \frac{2^x+1}{2^{x+1}}$$
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Hint:
$\displaystyle \frac{2^x + 1}{2^{x+1}} = \frac{2^x}{2^{x+1}} + \frac{1}{2^{x+1}} = \frac{1}{2} + \frac{1}{2^{x+1}}$
agha
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Darth Geek
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I see how that works you have to take out a 2 from $2^{x+1}$ so the $2x$ terms will cancel – user8028 Jul 22 '14 at 21:56
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I'm glad I could help =D – Darth Geek Jul 22 '14 at 22:03
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Hint: $\displaystyle\lim_{x \to \infty} \frac{2^x+1}{2^{x+1}} = \lim_{x \to \infty} \frac{2^x}{2^{x+1}}+\frac{1}{2^{x+1}}$ Can you simplify this?
JimmyK4542
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Write $2^{x+1}$ as ${2(2^x)}$, then use L'Hopital's rule. $$\lim_{x\to\infty}\frac{2^x\ln2}{2(2^x\ln2)}$$ Result is $\frac{1}{2}$.
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Wouldn't you still be left with an infinity over infinity situation because of the 2^x in the numerator and denominator? – user8028 Jul 22 '14 at 21:53
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Yes, but now, we can divide numberator and denominator by 2^x*ln(2). I think this is the easiest way. – Jul 22 '14 at 21:55
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Let $t=2^x$ therefore $t\to\infty \ \ \text{as} \ x\to\infty$ hence $$\lim\limits_{x\to\infty}\frac{2^x+1}{2 \cdot 2^x}=\lim\limits_{t\to\infty}\frac{t+1}{2t}=\frac{1}{2}.$$
user62498
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