Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

oh ok

Answer 5

and then from there which method would u use

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

hang on lolz

Answer 10

is it 0?

Answer 11

yup, its 0
good work :)

Answer 12

yay!!!

Answer 13

thank you!!

Answer 14

i need help again actually with number 84 now

Answer 15

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

Answer 16

im confused on tht part

Answer 17

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?

Answer 18

ok

Answer 19

and then what?

Answer 20

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 ?

Answer 21

yeah i got tht part

Answer 22

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 ?

Answer 23

oh so ur multiplying it by its conjugate right ?

Answer 24

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

Answer 25

oh ok

Answer 26

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

Answer 27

wait a minute now im confused a bit

Answer 28

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

Answer 29

oh ok

Answer 30

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 ?

Answer 31

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

Answer 32

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?

Answer 33

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

Answer 34

right?

Answer 35

idk lolz

Answer 36

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

Answer 37

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

Answer 38

is it?

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

oh ok

Answer 43

thanks!

Answer 44

welcome :)

Answer 45

u r a very helpful math teacher!