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Evaluate $$\lim_{x \to\infty}{\frac{1}{2^x-5^x+3^x}}. $$

My attempt:
$$\lim\limits_{x \rightarrow \infty}{\frac{1}{2^x-5^x+3^x}} = \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{1-\frac{5^x}{2^x}+\frac{3^x}{2^x}}} \\= \lim\limits_{x \rightarrow \infty}{\frac{1}{2^x} \cdot\frac{1}{\frac{1}{5^x}-\frac{5^x}{10^x}+\frac{3^x}{10^x}}\cdot\frac{1}{5^x}} = 0 \cdot\infty \cdot 0 = 0.$$
Someone told me that it is a wrong way to show the limit. Can anyone explain why?

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    $\begingroup$ Because $0\cdot\infty$ is an indeterminate form...? $\endgroup$ – CiaPan Aug 18 '17 at 7:47
  • $\begingroup$ You have problem when you go from limit expression to $0\cdot\infty\cdot 0$. This is not allowed via any theorem on limits. $\endgroup$ – Paramanand Singh Aug 18 '17 at 8:12
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Because $0\cdot \infty$ is an Indeterminate form. Note that $${\frac{1}{2^x-5^x+3^x}}=\left(\frac{1}{5}\right)^x\cdot \frac{1}{\left(\frac{2}{5}\right)^x-1+\left(\frac{3}{5}\right)^x}.$$ Since $1/5$, $2/5$, and $3/5$ are positive numbers less than $1$, what may we conclude as $x\to +\infty$?

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  • $\begingroup$ Obviously 0*(-1) = 0. I did not know about indetermined form yet neither I had the idea of pull the $5^x$ out of the fracture. Thanks for fixing my issue. $\endgroup$ – Maddude Aug 18 '17 at 7:59
  • $\begingroup$ @Maddude Yes, the idea here is to pull out the term that goes faster to infinity. $\endgroup$ – Robert Z Aug 18 '17 at 8:01
  • $\begingroup$ @user236182 Thanks for pointing out the typo! $\endgroup$ – Robert Z Aug 18 '17 at 10:44
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$0\cdot \infty$ is $inderterminate$. To see this, calculate the following limits: $$\lim_{x\to \infty} \frac{1}{x} \cdot x\qquad \text{and} \lim_{x\to \infty} 0\cdot x,$$ which are both of the form $0\cdot \infty$.

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Because it is not true that $0\times\infty\times0=0$. In fact, $0\times\infty\times0$ is indeterminate.

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  • $\begingroup$ Actually "$0\times\infty\times0$" is indeterminate, not "$0\times\infty\times0=0$". $\endgroup$ – CiaPan Aug 18 '17 at 7:51
  • $\begingroup$ @CiaPan Edited. Thanks. $\endgroup$ – José Carlos Santos Aug 18 '17 at 7:53

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