Question

Lim x to infinity x^(1/2)*sin(((x+2)^1/2) - ((x-2)^1/2))

Answer 1

is that:\[\sqrt{x}*\sin(\sqrt{x+2}-\sqrt{x-2})\]?

Answer 2

\[\lim_{x \rightarrow \infty} \sqrt{x} \sin (\sqrt{x+2}-\sqrt{x-2})\]

Answer 3

yeah

Answer 4

first simplify the term under sin.rationalise it

Answer 5

oops.sorry,that would complicate it further.no need to do that

Answer 6

why? thats a good point for start..

Answer 7

after doing that we just need to apply\[\sin z \approx z \]when z approaches to zero

Answer 8

if im not wrong :)

Answer 9

well the answer is 2 so idk

Answer 10

i just dont know how to get there

Answer 11

thats right answer is 2

Answer 12

ok see starting with the hint MotherofGod gave u\[\lim_{x \rightarrow \infty} \sqrt{x} \sin (\sqrt{x+2}-\sqrt{x-2})=\lim_{x \rightarrow \infty} \sqrt{x} \sin (\frac{4}{\sqrt{x+2}+\sqrt{x-2}})\]because\[\sqrt{x+2}-\sqrt{x-2}=\sqrt{x+2}-\sqrt{x-2}*\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-2}}=\frac{4}{\sqrt{x+2}+\sqrt{x-2}}\]

Answer 13

\[\frac{4}{\sqrt{x+2}+\sqrt{x-2}}\]goes to zero as x goes to infty so\[\sin(\frac{4}{\sqrt{x+2}+\sqrt{x-2}}) \approx \frac{4}{\sqrt{x+2}+\sqrt{x-2}}\]rewrite the limit\[\lim_{x \rightarrow \infty} \sqrt{x} \sin (\sqrt{x+2}-\sqrt{x-2})=\lim_{x \rightarrow \infty}\frac{4\sqrt{x}} {\sqrt{x+2}+\sqrt{x-2}}=\frac{4\sqrt{x}}{2\sqrt{x}}=2\]

Answer 14

(x+2)^1/2 + (x-2)^2 = 2x^1/2 ???

Answer 15

as x goes to infty 2 and -2 will be neglegible in comparison with x

Answer 16

yeahh!!!
Great!!! Thx a lot

Answer 17

very welcome :)