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November 18th, 2018, 08:28 PM   #1
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Hi everyone,
how can we take the limit of this function. I don't need the asnwer. I have it.
I just need to know the steps.

lim(x is app to infinity) (-1)^(x+1)*x/ x+(sqrtx)

Last edited by Leonardox; November 18th, 2018 at 08:36 PM.
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November 18th, 2018, 09:59 PM   #2
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$\lim \limits_{x \to \infty}~(-1)^{x+1}\cdot \dfrac{x}{x+\sqrt{x}}$

I don't believe this limit exists.

Divide top and bottom by $x$

$\lim \limits_{x \to \infty}~(-1)^{x+1}\cdot \dfrac{1}{1+\frac{1}{\sqrt{x}}}$

clearly $\lim \limits_{x\to\infty}~\dfrac{1}{\sqrt{x}} = 0$

so as $x \to \infty$ you start ping ponging back and forth between terms very close to 1 and terms very close to -1. This never converges, the two terms just get closer and closer to 1 and -1.
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November 18th, 2018, 11:34 PM   #3
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Also, if you write $x$, then to me that is a continuous variable, meaning all $x\in \mathbb{R}$ are allowed. In that case though, isn't $(-1)^{x+1}$ ill-defined?
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Last edited by greg1313; November 20th, 2018 at 01:28 AM.
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November 19th, 2018, 04:18 AM   #4
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For x-integer then (-1)^x is defined
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November 20th, 2018, 05:58 PM   #5
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$(-1)^x=e^{x\pi i}$
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