Question

Limit question

Answer 1

\[\LARGE \lim_{x \rightarrow a} \sin^{-1} \sqrt{\frac{a-x}{a+x}} \csc \sqrt{(a^2-x^2)}\]

Answer 2

\[\LARGE \lim_{x \rightarrow a} \sin^{-1} \sqrt{\frac{a-x}{a+x}} \frac{1}{\sin\sqrt{(a+x)}(\sqrt{a-x})}\]

Answer 3

It came in mains probably in 2014

Answer 4

no it didnt come..

Answer 5

ohk

Answer 6

im thinking of making a substitution \(\large x = a\cos (2\theta)\)

Answer 7

\[\large \lim_{x \rightarrow a} ~ \dfrac{\sin^{-1} \sqrt{\frac{a-x}{a+x}} }{\sin\sqrt{a^2-x^2}}\]

Answer 8

i was thinking the same but then what about the cosec part

Answer 9

substitute \(x = a\cos (2\theta)\)

as \(x \to a,~~ \theta \to \frac{1}{2}\arccos(\frac{x}{a})\)

Answer 10

\( \theta \to 0 \) i guess ^

Answer 11

yep 0

Answer 12

I feel x=acostheta would be a better option!

Answer 13

\[\large \lim_{\theta \rightarrow 0} ~ \dfrac{\sin^{-1} \sqrt{\frac{a-a\cos (2\theta)}{a+a\cos(2\theta)}} }{\sin\sqrt{a^2-(a\cos(2\theta))^2}}\]

Answer 14

hmm its the same thing :P okay

Answer 15

ok lets assume its acostheta

Answer 16

\[\large \lim_{\theta \rightarrow 0} ~ \dfrac{\sin^{-1} \tan \frac{\theta}{2} }{\sin asin \theta}\]

Answer 17

ah yeah theta/2 or theta lets assume cos 2theta only

Answer 18

your expression is correct...

Answer 19

lets stick to it : x = acostheta

Answer 20

\[\Huge \lim_{\theta \rightarrow 0} ~ \dfrac{\sin^{-1} \tan \frac{\theta}{2} }{\sin asin \theta}\]
assuming x=a cos theta

Answer 21

i would do lhosps next

Answer 22

\[\LARGE \frac{\frac{1}{1- \tan^2 \frac{\theta}{2}}}{\cos (a \sin \theta) \times a \cos \theta}\]

Answer 23

theta->0

Answer 24

\[\large \lim_{\theta \rightarrow 0} ~ \dfrac{\frac{1}{\sqrt{1-\tan^2(\frac{\theta}{2})}}\times \sec^2(\frac{\theta}{2} )\times \frac{1}{2}}{\cos(a\sin \theta)\times a\cos \theta } \]

Answer 25

take the limit

Answer 26

oh yeah hmm

Answer 27

1/2a :P

Answer 28

xD did u get why wolf is crying
http://www.wolframalpha.com/input/?i=%5Clim_%7Bx+%5Crightarrow+a%7D+arcsin%28%5Csqrt%7B%5Cfrac%7Ba-x%7D%7Ba%2Bx%7D%7D%29+%5Ccsc+%5Csqrt%7B%28a%5E2-x%5E2%29%7D

Answer 29

nope..:/

Answer 30

it got to do with domain of the function i guess

Answer 31

since the domain of arcsin(x) is between [-1, 1], the parameter \(a\) will be having some restrictions... not entire sure

Answer 32

oh...acha maybe