Question

check my math pls
lim as x ->infinite (x^2-x)/sqrt(x^4+x)

Answer 1

i have this so far \[(x^2/x^2-x/x^2)/\sqrt(x^4/x^4+x/x^4)\]

Answer 2

\[\lim_{x \rightarrow \infty}\frac{x^2-x}{\sqrt{x^4+x}}\]
highest power is inside the sqrt( )
divide both to and bottom by (sqrt(x^4))
which you just did

Answer 3

\[then (1-1/x)/\sqrt(1+1x^3)\]

Answer 4

\[then (1-0)/\sqrt(1+0)\]

Answer 5

and finally got 1

Answer 6

that is exactly right
\[\lim_{x \rightarrow \infty}\frac{\frac{x^2}{x^2}-\frac{x}{x^2}}{\sqrt{\frac{x^4}{x^4}+\frac{x}{x^4}}} =\lim_{x \rightarrow \infty} \frac{1-\frac{1}{x}}{\sqrt{1+\frac{1}{x^3}}}=\frac{1-0}{\sqrt{1+0}}=\frac{1}{\sqrt{1}}=\frac{1}{1}=1\]

Answer 7

yep :D

Answer 8

dont go yey i got a question

Answer 9

let i have the same equation but instead of x^4 i have x^9 as denominator

Answer 10

i would divide the top by x^3?

Answer 11

so if we have \[\lim_{x \rightarrow \infty}\frac{x^2-x}{\sqrt{x^9+x}}\]
divide top and bottom by sqrt(x^9)

Answer 12

by sqrt(x^9) doesn't equal x^3

Answer 13

what i meant was to divide the top by x^3

Answer 14

but the bottom by x^9

Answer 15

\[\lim_{x \rightarrow \infty}\frac{\frac{x^2}{x^\frac{9}{2}}-\frac{x}{ x^\frac{9}{2}}}{\sqrt{\frac{x^9}{x^9}+\frac{x}{x^9}}} =\lim_{x \rightarrow \infty}\frac{\frac{1}{x^\frac{5}{2}}-\frac{1}{x^\frac{7}{2}}}{ \sqrt{ 1-\frac{1}{x^\frac{7}{2}}}} \]

Answer 16

:O i see that 2 from x^9/2 is from the sqrt right? so if it was 3sqrt it would x^9/3 so just x^3

Answer 17

\[\lim_{x \rightarrow \infty}\frac{x^2-x}{\sqrt[3]{x^9-x}} =\lim_{x \rightarrow \infty}\frac{\frac{x^2}{x^3}-\frac{x}{x^3}}{\sqrt[3]{\frac{x^9}{x^9}-\frac{x}{x^9}}}\]
yea

Answer 18

cool thanks :D