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Please help me in explaining how to get the following results:

$\lim_{x \rightarrow 0} (x^{-2}-x^{-1}) = \infty$

and

$\lim_{x \rightarrow \infty} (3^x-2^x) = \infty$

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Thank you (Amzoti) for editing the question. I'm not sure how to activate the Latex arguments the posts –  Logarithm Mar 20 '13 at 5:25
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You can use L-hospital rule for the first limit and show that the limit tends to infinity. For the second limit you can use the $a^x$ expansion and then use limit on that series. –  lsp Mar 20 '13 at 5:29
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2 Answers

up vote 6 down vote accepted

Hints

$$==\;\;\;\;x^{-2}-x^{-1}=\frac{1-x}{x^2}\xrightarrow [x\to 0]{}\ldots$$

$$==\;\;\;\;3^x-2^x=3^x\left(1-\left(\frac{2}{3}\right)^x\right)\xrightarrow[x\to\infty]{}\ldots$$

Further hint:

$$|q|<1\Longrightarrow q^n\xrightarrow[n\to\infty]{}0$$

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Thank you so much! Mathematics is so beautiful ... –  Logarithm Mar 20 '13 at 5:37
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$\lim_{x \rightarrow \infty} 3^x(1-(\frac{2}{3})^x)$

now as ${x\rightarrow\infty}$

$(\frac{2}{3})^x\rightarrow0$

and

$3^{x}\rightarrow \infty$

hence the final answer is $\infty$

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