Question

Limit
http://puu.sh/d37vc/fc565b4d05.png

Answer 1

\[\lim_{x \rightarrow 1}\frac{\sqrt{x}-x^2}{1-\sqrt{x}}\]

Answer 2

try a substitution maybe
let sqrt(x)=u

Answer 3

havent learned how to :P

Answer 4

ok maybe try factoring on a sqrt(x) on top

Answer 5

and then you will factor a difference of cubes on top

Answer 6

then you will cancel a common factor on top and bottom

Answer 7

\[\sqrt{x}-x^2=\sqrt{x}(1-(\sqrt{x})^3)=..\]
use the difference of cubes formula there

Answer 8

you will see that we can cancel the 1-sqrt(x) on top and bottom

Answer 9

\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 10

that is the formula (or identity or whatever you formally call it) that I'm asking you to apply to the 1-(sqrt(x))^3

Answer 11

ah i see and then apply 1 into x?

Answer 12

after canceling that bottom out with some common factor on top yes

Answer 13

ah ok thank you

Answer 14

did you factored that one part?

Answer 15

did you want to see if you got the correct factored form

Answer 16

or the right limit

Answer 17

here i will let you check your factored form
\[a^3-b^3=(a-b)(a^2+ab+b^2) \\ 1-(\sqrt{x})^3=(1-\sqrt{x})(1+\sqrt{x}+x) \\ \text{ so } \sqrt{x}(1-(\sqrt{x})^3)=\sqrt{x}(1-\sqrt{x})(1+\sqrt{x}+x)\]
that is what the top should look like after factoring it such that the bottom and top have an obvious common factor

Answer 18

if you need further assistance
i should brb

Answer 19

and the answer should be 3 then?

Answer 20

yes that is correct

Answer 21

thank you