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anonymous
 4 years ago
can someone help my find the limit if x>0 sinx/ cubed root of x
anonymous
 4 years ago
can someone help my find the limit if x>0 sinx/ cubed root of x

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hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1multiply and divide by x , to get the function in the form of sin x/ x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can u draw it out so i can see what ur talking about

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1\(\huge\frac{sin x}{x}\:\frac{x}{\sqrt[3]{x}}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and then from there which method would u use

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1u know the formula : \[\lim_{x \rightarrow 0}\: sinx / x = 1\] ? use that for first fraction.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh ok! omg now it makes sense thank you @hartnn

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1welcome :) what did u get the final answer as ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need help again actually with number 84 now

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1ok, if f(x) = root x what will be f(x+h) ?? u know?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im confused on tht part

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1ok, to get f(x+h) just replace x by (x+h) in f(x) so f(x+h) will be \(\sqrt{x+h}\) ok?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1i 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 ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1now 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 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh so ur multiplying it by its conjugate right ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yup, so that i get only h in numerator and that i can cancel it with h in denominator ok?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1i made a typo ..... it should be \(\sqrt{x} \) and not \(\sqrt{h} \) in the numerator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait a minute now im confused a bit

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1because itsf(x+h)  f(x) which is \(\sqrt{x+h} \) \(\sqrt{x} \)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1so 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 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would it still be h in the numerator i think..

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yup, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you have \[\sqrt{x+h} + \sqrt{x}\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yes, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x+0} +\sqrt{x}\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yup, thats your denominator = \(2\sqrt{x}\) and then your final answer would be \(\huge\frac{1}{2\sqrt{x}}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hang on im confused how do u get \[\frac{ 1 }{ 2\sqrt{x} }\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1\(\sqrt{x+0} +\sqrt{x}=\sqrt{x} +\sqrt{x}=2\sqrt{x}\) and there was 1 in the numerator, previously, right ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u r a very helpful math teacher!
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