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## anonymous 3 years ago can someone help my find the limit if x->0 sinx/ cubed root of x

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1. hartnn

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

2. anonymous

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

3. hartnn

$$\huge\frac{sin x}{x}\:\frac{x}{\sqrt[3]{x}}$$

4. anonymous

oh ok

5. anonymous

and then from there which method would u use

6. hartnn

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

7. anonymous

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

8. hartnn

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

9. anonymous

hang on lolz

10. anonymous

is it 0?

11. hartnn

yup, its 0 good work :)

12. anonymous

yay!!!

13. anonymous

thank you!!

14. anonymous

i need help again actually with number 84 now

15. hartnn

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

16. anonymous

im confused on tht part

17. hartnn

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?

18. anonymous

ok

19. anonymous

and then what?

20. hartnn

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 ?

21. anonymous

yeah i got tht part

22. hartnn

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 ?

23. anonymous

oh so ur multiplying it by its conjugate right ?

24. hartnn

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

25. anonymous

oh ok

26. hartnn

i made a typo ..... it should be $$-\sqrt{x}$$ and not $$-\sqrt{h}$$ in the numerator

27. anonymous

wait a minute now im confused a bit

28. hartnn

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

29. anonymous

oh ok

30. hartnn

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 ?

31. anonymous

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

32. hartnn

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?

33. anonymous

do you have $\sqrt{x+h} + \sqrt{x}$

34. anonymous

right?

35. anonymous

idk lolz

36. hartnn

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

37. anonymous

$\sqrt{x+0} +\sqrt{x}$

38. anonymous

is it?

39. hartnn

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

40. anonymous

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

41. hartnn

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

42. anonymous

oh ok

43. anonymous

thanks!

44. hartnn

welcome :)

45. anonymous

u r a very helpful math teacher!

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