alexthomas
Find the limit if it exists...
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alexthomas
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|dw:1372210588807:dw|
alexthomas
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the radical is over the (+x) as well if not clear
Jhannybean
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You want to multiply by the conjugate
Jhannybean
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\[\large \lim_{x \rightarrow \infty} \ \frac{\sqrt{25x^2+x}-5x}{1} \cdot \frac{\sqrt{25x^2+x} +5x}{\sqrt{25x^2+x} +5x}\]
alexthomas
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i'm getting 25+x-25x^2 /radical 25x^2+x +5x
Jhannybean
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\[\large \lim_{x \rightarrow \infty} \ \frac{(25x^2+x)-(5x)^2}{\sqrt{25x^2+x} +5x} \]
alexthomas
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exactly
Jhannybean
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mmhmm.
Jhannybean
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\[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x}\]
alexthomas
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good i got that too it cant be simplified any further right?
dumbcow
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i would use l'hopitals rule now
Jhannybean
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Ah. Okay.
Jhannybean
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LH Rule: \[\large \frac{ f'(x)}{g'(x)}\]
Jhannybean
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Since if we took the limit to infinity we'd get \(\large \frac{\infty}{\infty}\)
Jhannybean
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Can you tell me what the derivative of \(\large \sqrt{25x^2 +x} +5x\) is?
alexthomas
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|dw:1372212305372:dw|
Jhannybean
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Not exactly.. We have \(\large \sqrt{25x^2+x}+5x\) right?
That can be rewritten as \(\large (25x^2 +x)^{1/2}+5x\)
Using chain rule: \(\large \frac{1}{2\sqrt{25x^2+x}} \cdot (50x +1) +5\)
Simplify : \(\large \frac{50x+1}{2\sqrt{25x^2+x}} +5\)
Jhannybean
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So now we'll have... \[\huge \lim_{x \rightarrow \infty} \ \frac{\frac{1}{50x+1}}{2\sqrt{25x^2+x}+5}\]
alexthomas
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i got the answer to be 1/10 thank you
Jhannybean
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Can you explain to me how you got that? :o
Jhannybean
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what i got was \[\large \lim_{x \rightarrow \infty} \ \frac{1}{(100x+2)(\sqrt{25x^2+x}+5)} = 0 \]
Jhannybean
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An unforseen method of approach!
dan815
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|dw:1372214578635:dw|
dan815
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what shud we subtract there
Jhannybean
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Uhhh...
dan815
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i dunno for some reason i feel like it cant be 0
dan815
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|dw:1372215149908:dw|
Jhannybean
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Hmmm...
dan815
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1/10 is probably right
Jhannybean
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How do you factor out a -5x from \(\large \sqrt{25x^2+x}\) ?
dan815
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its tough sqrt roots arent friendly
Jhannybean
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HOW DO YOU GET 1/10???? thats what i wanted to know! and then the guy disappeared lol.
dan815
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gotta use other ways
dan815
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i didnt im just saying that answer looks right lol
Jhannybean
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secret methods....hidden treasures.
Jhannybean
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I want toknow this.
dan815
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maybe we can rewrite this as some other known function base like hyperbola?
RaphaelFilgueiras
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|dw:1372215364565:dw|
dan815
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oh ok
Jhannybean
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what??...
RaphaelFilgueiras
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dont need to use L'H,just put x in evidence
Jhannybean
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im so confused.....wait a minute...
dan815
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shudda known it was something simple
dan815
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basically hes saying
the highest exponents there will just be root(25x^2) and 5x so we can still divide everything by x
Jhannybean
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Oh...hm. Let me try this.
RaphaelFilgueiras
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|dw:1372215619036:dw|
dan815
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|dw:1372215830146:dw|
dan815
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@whpalmer4 show us the way
RaphaelFilgueiras
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|dw:1372215784059:dw|
Jhannybean
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Am i wrong?
RaphaelFilgueiras
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i think that is not right
dan815
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u shud have +5 there instead of last 0, and ur diving by x that is the higest power
Jhannybean
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Oh YEAH. 5x/x * my bad.
Jhannybean
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I forgot the x......
1/10 is the answer,lol. whoops.
RaphaelFilgueiras
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@dan815 x²(25+1/x)= 25x²+x
dan815
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|dw:1372216059586:dw|
whpalmer4
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you were on the right track
\[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x} = \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}}\]
RaphaelFilgueiras
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|dw:1372216149948:dw|
Jhannybean
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\[\large \lim_{x \rightarrow \infty} \ \frac{x/x}{\sqrt{25x^2/x^2 +x/x^2}+5x/x}\]\[\large \frac{1}{\sqrt{25+0}+5} =\frac{1}{10} \]
Jhannybean
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When lim tends to infinity, divide by highest exponent. When lim tends to a number, use LH rule :D
RaphaelFilgueiras
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just take care because if x goes to - inf |x|=-x
Jhannybean
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yeah. I understand :)
RaphaelFilgueiras
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sqrt(x²)=|x| not x; we do that only because we know x is positive(inf)
Jhannybean
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You divide everything under a square root by the highest power underneath the root, and all the other x's get divided by the value of x that is dependent on the behavior of the function, (i.e if x tends to positive or negative infinity)
Jhannybean
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lol we're saying the same thing....2 different ways :(
RaphaelFilgueiras
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if x goes to - inf the result isn't the same
Jhannybean
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yes,so if x goes to -inf, x takes on neg values.... hence we divide by -x instead of x.
whpalmer4
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\[\large \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}} = \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{25x^2+x}{x^2}}} \]\[= \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{50x+1}{2x}}}\]by L'Hopital
\[=\frac{1}{5+\sqrt{25}}\]
Jhannybean
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Oh, I see.
Jhannybean
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Tried @whpalmer4 's way myself just now. REALLY interesting hahaha