Question

Pretty hard limit problem

Answer 1

\[\lim_{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{1-\cos \sqrt{x}}\]

Answer 2

@mukushla hi

Answer 3

hi :)

Answer 4

i'd say rationalize both num and denum and use the formula:\[1-\cos^2 \theta=\sin^2 \theta\]

Answer 5

@mukushla and how could I rationalize the denominator?

Answer 6

are you supposed to be doing this algebraically?

Answer 7

yes

Answer 8

have you figured it out yet?

Answer 9

Nope

Answer 10

start by multiplying top and bottom by

\[1+\sqrt{\cos x}\]

and

\[1+\cos \sqrt{x}\]

Answer 11

I get
\[\frac{ (1-\cos x)(1 + \cos \sqrt{x}) }{ (1+\sqrt{\cos x})(1-\cos^2\sqrt{x}) }\]

Answer 12

ok...rewite \[1-\cos^2(\sqrt{x})\] in terms of sine


then multiply top and bottom by \(1+\cos(x)\)

Answer 13

ok, I got
\[\frac{(1-\cos^2x)(1+\cos\sqrt{x})}{(1+\sqrt{\cos x})(1+\cos x)\sin^2\sqrt{x}}\]

Answer 14

and \(1-\cos^2(x)=\sin^2(x)\)

now multiply top and bottom by x and multiply top and bottom by \(\sqrt{x}\)

and use the fact that \[\frac{\sin(u)}{u}\to1\text{ as }u\to 0\]

Answer 15

sorry ...multiply top and bottom by x^2 and \(\sqrt{x}^2=x\)

Answer 16

you get it?

Answer 17

Yes, I think so. Limit is equal to 0.5?
O don't really understand why the limit of \[\frac{ x }{ \sin^2x }\] as x approaches 0 is 1. How do you get it form the previous limit?

Answer 18

no the limit is 0

Answer 19

yes, you are right. I Made a mistake

Answer 20

\[\frac{(1-\cos^2x)(1+\cos\sqrt{x})}{(1+\sqrt{\cos x})(1+\cos x)\sin^2\sqrt{x}}\]

\[=\frac{\sin^2(x)(1+\cos\sqrt{x})}{(1+\sqrt{\cos x})(1+\cos x)\sin^2\sqrt{x}}\]

\[=\frac{(1+\cos\sqrt{x})}{(1+\sqrt{\cos x})(1+\cos x)}\frac{\sin^2(x)}{1}\frac{1}{\sin^2(\sqrt{x})}\]

\[=\frac{(1+\cos\sqrt{x})}{(1+\sqrt{\cos x})(1+\cos x)}\frac{\sin^2(x)}{x^2}\frac{x}{\sin^2(\sqrt{x})}\frac{x^2}{x}\]

\[=\frac{(1+\cos\sqrt{x})}{(1+\sqrt{\cos x})(1+\cos x)}\frac{\sin^2(x)}{x^2}\frac{x}{\sin^2(\sqrt{x})}x\]

as \(x\to 0\)

\[\frac{2}{2\cdot 2}\cdot 1\cdot 1\cdot 0=0\]