Question

Could somebody help e with this limit ? D:
lim 1-cosx/x^2/3 approaches x-> 0 please:D

Answer 1

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x^\frac{2}{3}}\]
is this right?

Answer 2

if so try multiplying top and bottom by x^(1/3)

Answer 3

you should know a famous trig limit you can use

Answer 4

dang can't figure out how to typeset l\Hopital
not that it is needed...

Answer 5

yeah is the first one ! But I can't Im new in this topic

Answer 6

do you know what this equals:
\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x}=?\]

Answer 7

zero right?

Answer 8

yes

Answer 9

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x^\frac{2}{3}} \cdot \frac{x^\frac{1}{3}}{x^\frac{1}{3}}\]
can you figure out why I want to multiply top and bottom by x^(1/3)?

Answer 10

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x} \cdot x^\frac{1}{3}\]

Answer 11

you can write this as
\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x} \cdot \lim_{x \rightarrow 0}x^\frac{1}{3}\]
since both limits exist

Answer 12

you already told me the first limit in the product equal 0
the second function is continuous at x=0 so you can use direct sub for the second limit in the product

Answer 13

so the final answer will be 0 because if I multiply \[x^{1/3} \] in 1-cos x , that will be zero and the denominator will be x so when i evaluate it will be 0/ 1=0 right?

Answer 14

well (1-cos(x))/x->0
and so you have 0*0
which is 0

Answer 15

Thanks finally I understood :D