Question

\[\lim_{R\to+\infty}\frac{x^2}{\sqrt{R^2+x^2}-R}%&=\\%&=\]

Answer 1

multiply with conjucate ?

Answer 2

you would got limit \( (\sqrt( R^2+x^2) +R \) which is \(\infty \)

Answer 3

\[
\lim_{R\to+\infty}\frac{x^2}{\sqrt{R^2+x^2}-R}\\
=\lim_{R\to+\infty}\frac{x^2}{\sqrt{R^2+x^2}-R}\times\frac{\sqrt{R^2+x^2}+R}{\sqrt{R^2+x^2}+R}\\
=\lim_{R\to+\infty}\frac{x^2(\sqrt{R^2+x^2}+R)}{(R^2+x^2)-R^2}\\
=
\]

Answer 4

\[=\lim_{R\to+\infty}\frac{x^2(\sqrt{R^2+x^2}+R)}{x^2}\\
=\lim_{R\to+\infty}\sqrt{R^2+x^2}+R \]

Answer 5

which should just yield \(\infty\)

Answer 6

hmm, it was meant to become zero,

Answer 7

whops, i meant

\[\begin{align}
&= \lim_{R\to+\infty}\frac{x^2}{\sqrt{R^2+x^2}+R} \\
&= \lim_{R\to+\infty}\frac{x^2}{\sqrt{R^2+x^2}+R}
\times\frac{\sqrt{R^2+x^2}-R}{\sqrt{R^2+x^2}-R} \\
&=\lim_{R\to+\infty}\frac{x^2(\sqrt{R^2+x^2}-R)}{(R^2+x^2)-R^2} \\
&=\lim_{R\to+\infty}\frac{x^2(\sqrt{R^2+x^2}-R)}{x^2} \\
&=\lim_{R\to+\infty}\sqrt{R^2+x^2}-R %&\leadsto\infty \\
%&
%&\neq 0
\end{align}\]

Answer 8

How can i show this one equals zero

Answer 9

try factoring R^2?

Answer 10

then factor R

Answer 11

you will get -1x0

Answer 12

Well when you got
\[ \lim_{R\to+\infty}\frac{x^2}{\sqrt{R^2+x^2}+R}\]
You can plug in infnity directly in the denominatory and you get \( \infty + \infty\) which is okay and defined to be \(\infty\), so you essentially have \[ \lim_{R \rightarrow \infty}\frac{x^2}{R}=0\], it behaves like 1/x as x goes to infinity

Answer 13

wait this looks like your original problem but but the denominator changes signs.. hm I don't think I read your full solution correctly xD

Answer 14

yeah actually from the start it is zero no further steps needed

Answer 15

hmm ok , thanks

Answer 16

well from the very start you have \(\infty - \infty\) in the denominator which is undefined

Answer 17

no he said that is + not negative
so it like you wrote x^2/R which tends to 0

Answer 18

i see now that sign in the denominator changes everything,

Answer 19

yeah any sign would change the things lol

Answer 20

or so it iis + in the denominator, then that is great! :)

Answer 21

@ikram002p, @kirbykirby , @xapproachesinfinity

thanks y'all

Answer 22

:)

Answer 23

yw!

Answer 24

(i was just trying to understand physics lecture notes, and i had copied a sign wrong, but you've all cleared it up now)