WEBVTT
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okay. We need to look at the Limited and
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goes to infinity of a m. Figure out whether
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or not that limit exists and whether or not it's
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fine. So this is limit as n goes to
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infinity of in squared over square root of in cubed
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plus foreign. This problem we do a similar trick
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to what we've done before As far as we look
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in the denominator, we looked to see what term
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in the denominator is going to infinity the fastest except
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here instead of dividing top and bottom by that term
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will need to factor out something. So we're gonna
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need to factor out this in cubed that we have
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happening here. So factor that out in the denominator
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and the square root function is multiplication. So we
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just gotta pull that outside like that. And then
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we have one plus for over and squared here.
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So now we can just rewrite this a little bit
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. Mhm in squared square root of n cubed is
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in to the 3/2. Mhm. Mhm. Okay
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, and now we can simplify this part a bit
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. So this is gonna be in to the Tu
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minus 3/2 so into the one half. And then
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we're still dividing by square root of one plus four
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over n squared. Okay, so hopefully that's legible
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. Okay, so now if we do that limit
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on top, divided by limit on bottom, which
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is something that we're allowed to do as long as
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we don't get indeterminate form, then we'll see that
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we're going to get infinity divided by one, which
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is just gonna be infinity. Yeah, right.
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So we're allowed to limit on top over limit on
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bottom as long as we don't get indeterminate form.
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Indeterminate form would be like infinity over infinity or something
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divided by zero. If we try doing that here
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, we're just gonna get infinity over one which is
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not considered indeterminate form. So that's fine. Okay
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, So the limit turns out to be infinity,
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so the limit exists, but it's not finite.
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So the sequence is still said to diverge