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alexthomasBest ResponseYou've already chosen the best response.0
dw:1372210588807:dw
 9 months ago

alexthomasBest ResponseYou've already chosen the best response.0
the radical is over the (+x) as well if not clear
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
You want to multiply by the conjugate
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
\[\large \lim_{x \rightarrow \infty} \ \frac{\sqrt{25x^2+x}5x}{1} \cdot \frac{\sqrt{25x^2+x} +5x}{\sqrt{25x^2+x} +5x}\]
 9 months ago

alexthomasBest ResponseYou've already chosen the best response.0
i'm getting 25+x25x^2 /radical 25x^2+x +5x
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
\[\large \lim_{x \rightarrow \infty} \ \frac{(25x^2+x)(5x)^2}{\sqrt{25x^2+x} +5x} \]
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
\[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x}\]
 9 months ago

alexthomasBest ResponseYou've already chosen the best response.0
good i got that too it cant be simplified any further right?
 9 months ago

dumbcowBest ResponseYou've already chosen the best response.0
i would use l'hopitals rule now
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
LH Rule: \[\large \frac{ f'(x)}{g'(x)}\]
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Since if we took the limit to infinity we'd get \(\large \frac{\infty}{\infty}\)
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Can you tell me what the derivative of \(\large \sqrt{25x^2 +x} +5x\) is?
 9 months ago

alexthomasBest ResponseYou've already chosen the best response.0
dw:1372212305372:dw
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
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 : \(\large \frac{50x+1}{2\sqrt{25x^2+x}} +5\)
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
So now we'll have... \[\huge \lim_{x \rightarrow \infty} \ \frac{\frac{1}{50x+1}}{2\sqrt{25x^2+x}+5}\]
 9 months ago

alexthomasBest ResponseYou've already chosen the best response.0
i got the answer to be 1/10 thank you
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Can you explain to me how you got that? :o
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
what i got was \[\large \lim_{x \rightarrow \infty} \ \frac{1}{(100x+2)(\sqrt{25x^2+x}+5)} = 0 \]
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
An unforseen method of approach!
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
what shud we subtract there
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
i dunno for some reason i feel like it cant be 0
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
How do you factor out a 5x from \(\large \sqrt{25x^2+x}\) ?
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
its tough sqrt roots arent friendly
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
HOW DO YOU GET 1/10???? thats what i wanted to know! and then the guy disappeared lol.
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
i didnt im just saying that answer looks right lol
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
secret methods....hidden treasures.
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
I want toknow this.
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
maybe we can rewrite this as some other known function base like hyperbola?
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
dw:1372215364565:dw
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
dont need to use L'H,just put x in evidence
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
im so confused.....wait a minute...
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
shudda known it was something simple
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
basically hes saying the highest exponents there will just be root(25x^2) and 5x so we can still divide everything by x
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Oh...hm. Let me try this.
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
dw:1372215619036:dw
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
@whpalmer4 show us the way
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
dw:1372215784059:dw
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
i think that is not right
 9 months ago

dan815Best ResponseYou've already chosen the best response.0
u shud have +5 there instead of last 0, and ur diving by x that is the higest power
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Oh YEAH. 5x/x * my bad.
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
I forgot the x...... 1/10 is the answer,lol. whoops.
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
@dan815 x²(25+1/x)= 25x²+x
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
you were on the right track \[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x} = \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}}\]
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
dw:1372216149948:dw
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
\[\large \lim_{x \rightarrow \infty} \ \frac{x/x}{\sqrt{25x^2/x^2 +x/x^2}+5x/x}\]\[\large \frac{1}{\sqrt{25+0}+5} =\frac{1}{10} \]
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
When lim tends to infinity, divide by highest exponent. When lim tends to a number, use LH rule :D
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
just take care because if x goes to  inf x=x
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
yeah. I understand :)
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
sqrt(x²)=x not x; we do that only because we know x is positive(inf)
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
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)
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
lol we're saying the same thing....2 different ways :(
 9 months ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.2
if x goes to  inf the result isn't the same
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
yes,so if x goes to inf, x takes on neg values.... hence we divide by x instead of x.
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[\large \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}} = \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{25x^2+x}{x^2}}} \]\[= \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{50x+1}{2x}}}\]by L'Hopital \[=\frac{1}{5+\sqrt{25}}\]
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.1
Tried @whpalmer4 's way myself just now. REALLY interesting hahaha
 9 months ago
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