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anonymous
 3 years ago
Find the limit if it exists...
anonymous
 3 years ago
Find the limit if it exists...

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372210588807:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the radical is over the (+x) as well if not clear

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1You want to multiply by the conjugate

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'm getting 25+x25x^2 /radical 25x^2+x +5x

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good i got that too it cant be simplified any further right?

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.0i would use l'hopitals rule now

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1LH Rule: \[\large \frac{ f'(x)}{g'(x)}\]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372212305372:dw

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1Not 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
 3 years ago
Best ResponseYou've already chosen the best response.1So now we'll have... \[\huge \lim_{x \rightarrow \infty} \ \frac{\frac{1}{50x+1}}{2\sqrt{25x^2+x}+5}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got the answer to be 1/10 thank you

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1Can you explain to me how you got that? :o

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

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1An unforseen method of approach!

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0what shud we subtract there

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0i dunno for some reason i feel like it cant be 0

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

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0its tough sqrt roots arent friendly

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

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0i didnt im just saying that answer looks right lol

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1secret methods....hidden treasures.

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0maybe we can rewrite this as some other known function base like hyperbola?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372215364565:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dont need to use L'H,just put x in evidence

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1im so confused.....wait a minute...

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0shudda known it was something simple

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0basically hes saying the highest exponents there will just be root(25x^2) and 5x so we can still divide everything by x

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1Oh...hm. Let me try this.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372215619036:dw

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0@whpalmer4 show us the way

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372215784059:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think that is not right

dan815
 3 years ago
Best ResponseYou've already chosen the best response.0u shud have +5 there instead of last 0, and ur diving by x that is the higest power

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1Oh YEAH. 5x/x * my bad.

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1I forgot the x...... 1/10 is the answer,lol. whoops.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@dan815 x²(25+1/x)= 25x²+x

whpalmer4
 3 years ago
Best ResponseYou've already chosen the best response.1you were on the right track \[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x} = \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1372216149948:dw

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just take care because if x goes to  inf x=x

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1yeah. I understand :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sqrt(x²)=x not x; we do that only because we know x is positive(inf)

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1You 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
 3 years ago
Best ResponseYou've already chosen the best response.1lol we're saying the same thing....2 different ways :(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if x goes to  inf the result isn't the same

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

whpalmer4
 3 years ago
Best 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 \[=\frac{1}{5+\sqrt{25}}\]

Jhannybean
 3 years ago
Best ResponseYou've already chosen the best response.1Tried @whpalmer4 's way myself just now. REALLY interesting hahaha
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