abannavong
can someone help my find the limit if x>0 sinx/ cubed root of x



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hartnn
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multiply and divide by x , to get the function in the form of sin x/ x

abannavong
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can u draw it out so i can see what ur talking about

hartnn
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\(\huge\frac{sin x}{x}\:\frac{x}{\sqrt[3]{x}}\)

abannavong
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oh ok

abannavong
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and then from there which method would u use

hartnn
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u know the formula :
\[\lim_{x \rightarrow 0}\: sinx / x = 1\]
?
use that for first fraction.

abannavong
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oh ok! omg now it makes sense thank you @hartnn

hartnn
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welcome :)
what did u get the final answer as ?

abannavong
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hang on lolz

abannavong
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is it 0?

hartnn
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yup, its 0
good work :)

abannavong
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yay!!!

abannavong
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thank you!!

abannavong
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i need help again actually with number 84 now

hartnn
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ok, if f(x) = root x
what will be f(x+h) ??
u know?

abannavong
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im confused on tht part

hartnn
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ok, to get f(x+h) just replace x by (x+h) in f(x)
so
f(x+h) will be \(\sqrt{x+h}\)
ok?

abannavong
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ok

abannavong
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and then what?

hartnn
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i m using h instead of delta x just for convenience here.
so now u put that in {f(x+h)f(x) }/ h
to get
\(\frac{\sqrt{x+h}\sqrt{h}}{h}\)
got this ?

abannavong
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yeah i got tht part

hartnn
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now mutiply and divide by
\(\sqrt{x+h}+\sqrt{h}\)
and use (a+b)(ab) = a^2  b^2
and tell me what u get in numerator ?

abannavong
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oh so ur multiplying it by its conjugate right ?

hartnn
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yup, so that i get only h in numerator and that i can cancel it with h in denominator
ok?

abannavong
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oh ok

hartnn
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i made a typo ..... it should be \(\sqrt{x} \) and not \(\sqrt{h} \)
in the numerator

abannavong
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wait a minute now im confused a bit

hartnn
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because itsf(x+h)  f(x) which is \(\sqrt{x+h} \) \(\sqrt{x} \)

abannavong
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oh ok

hartnn
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so now now mutiply and divide by
\(\sqrt{x+h}+\sqrt{x}\)
and use (a+b)(ab) = a^2  b^2
and tell me what u get in numerator ?

abannavong
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would it still be h in the numerator i think..

hartnn
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yup, because (x+h) x = x+hx = h
this h cancels out with h in denominator and u have only 1 in numerator, right ?
and what u have in denominator now?

abannavong
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do you have \[\sqrt{x+h} + \sqrt{x}\]

abannavong
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right?

abannavong
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idk lolz

hartnn
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yes, u are right, be confident :)
now just put h =0
because lim h> 0
and tell me what u get in denominator after putting h=0

abannavong
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\[\sqrt{x+0} +\sqrt{x}\]

abannavong
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is it?

hartnn
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yup, thats your denominator = \(2\sqrt{x}\)
and then your final answer would be
\(\huge\frac{1}{2\sqrt{x}}\)

abannavong
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hang on im confused how do u get \[\frac{ 1 }{ 2\sqrt{x} }\]

hartnn
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\(\sqrt{x+0} +\sqrt{x}=\sqrt{x} +\sqrt{x}=2\sqrt{x}\)
and there was 1 in the numerator, previously, right ?

abannavong
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oh ok

abannavong
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thanks!

hartnn
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welcome :)

abannavong
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u r a very helpful math teacher!