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|dw:1372210588807:dw|

the radical is over the (+x) as well if not clear

You want to multiply by the conjugate

i'm getting 25+x-25x^2 /radical 25x^2+x +5x

\[\large \lim_{x \rightarrow \infty} \ \frac{(25x^2+x)-(5x)^2}{\sqrt{25x^2+x} +5x} \]

exactly

mmhmm.

\[\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x}\]

good i got that too it cant be simplified any further right?

i would use l'hopitals rule now

Ah. Okay.

LH Rule: \[\large \frac{ f'(x)}{g'(x)}\]

Since if we took the limit to infinity we'd get \(\large \frac{\infty}{\infty}\)

Can you tell me what the derivative of \(\large \sqrt{25x^2 +x} +5x\) is?

|dw:1372212305372:dw|

i got the answer to be 1/10 thank you

Can you explain to me how you got that? :o

what i got was \[\large \lim_{x \rightarrow \infty} \ \frac{1}{(100x+2)(\sqrt{25x^2+x}+5)} = 0 \]

An unforseen method of approach!

|dw:1372214578635:dw|

what shud we subtract there

Uhhh...

i dunno for some reason i feel like it cant be 0

|dw:1372215149908:dw|

Hmmm...

1/10 is probably right

How do you factor out a -5x from \(\large \sqrt{25x^2+x}\) ?

its tough sqrt roots arent friendly

HOW DO YOU GET 1/10???? thats what i wanted to know! and then the guy disappeared lol.

gotta use other ways

i didnt im just saying that answer looks right lol

secret methods....hidden treasures.

I want toknow this.

maybe we can rewrite this as some other known function base like hyperbola?

|dw:1372215364565:dw|

oh ok

what??...

dont need to use L'H,just put x in evidence

im so confused.....wait a minute...

shudda known it was something simple

Oh...hm. Let me try this.

|dw:1372215619036:dw|

|dw:1372215830146:dw|

@whpalmer4 show us the way

|dw:1372215784059:dw|

Am i wrong?

i think that is not right

u shud have +5 there instead of last 0, and ur diving by x that is the higest power

Oh YEAH. 5x/x * my bad.

I forgot the x......
1/10 is the answer,lol. whoops.

|dw:1372216059586:dw|

|dw:1372216149948:dw|

When lim tends to infinity, divide by highest exponent. When lim tends to a number, use LH rule :D

just take care because if x goes to - inf |x|=-x

yeah. I understand :)

sqrt(x²)=|x| not x; we do that only because we know x is positive(inf)

lol we're saying the same thing....2 different ways :(

if x goes to - inf the result isn't the same

yes,so if x goes to -inf, x takes on neg values.... hence we divide by -x instead of x.

Oh, I see.

Tried @whpalmer4 's way myself just now. REALLY interesting hahaha