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

Mathematics
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multiply and divide by x , to get the function in the form of sin x/ x
can u draw it out so i can see what ur talking about
\(\huge\frac{sin x}{x}\:\frac{x}{\sqrt[3]{x}}\)

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Other answers:

oh ok
and then from there which method would u use
u know the formula : \[\lim_{x \rightarrow 0}\: sinx / x = 1\] ? use that for first fraction.
oh ok! omg now it makes sense thank you @hartnn
welcome :) what did u get the final answer as ?
hang on lolz
is it 0?
yup, its 0 good work :)
yay!!!
thank you!!
i need help again actually with number 84 now
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ok, if f(x) = root x what will be f(x+h) ?? u know?
im confused on tht part
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?
ok
and then what?
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 ?
yeah i got tht part
now mutiply and divide by \(\sqrt{x+h}+\sqrt{h}\) and use (a+b)(a-b) = a^2 - b^2 and tell me what u get in numerator ?
oh so ur multiplying it by its conjugate right ?
yup, so that i get only h in numerator and that i can cancel it with h in denominator ok?
oh ok
i made a typo ..... it should be \(-\sqrt{x} \) and not \(-\sqrt{h} \) in the numerator
wait a minute now im confused a bit
because itsf(x+h) - f(x) which is \(\sqrt{x+h} \) \(-\sqrt{x} \)
oh ok
so now now mutiply and divide by \(\sqrt{x+h}+\sqrt{x}\) and use (a+b)(a-b) = a^2 - b^2 and tell me what u get in numerator ?
would it still be h in the numerator i think..
yup, because (x+h)- x = x+h-x = h this h cancels out with h in denominator and u have only 1 in numerator, right ? and what u have in denominator now?
do you have \[\sqrt{x+h} + \sqrt{x}\]
right?
idk lolz
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
\[\sqrt{x+0} +\sqrt{x}\]
is it?
yup, thats your denominator = \(2\sqrt{x}\) and then your final answer would be \(\huge\frac{1}{2\sqrt{x}}\)
hang on im confused how do u get \[\frac{ 1 }{ 2\sqrt{x} }\]
\(\sqrt{x+0} +\sqrt{x}=\sqrt{x} +\sqrt{x}=2\sqrt{x}\) and there was 1 in the numerator, previously, right ?
oh ok
thanks!
welcome :)
u r a very helpful math teacher!

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