Question

Evaluate a limit:
limx→0 x
−−−−−−−-
√1−cosx

Answer 1

x over sqrt of 1-cosx

Answer 2

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

Answer 3

\[\LARGE \lim_{x \rightarrow 0} \frac{x}{\sqrt{1-\cos x}} \times \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}=?\]

Answer 4

I don't know the numerator. Could you teach me:)

Answer 5

\[\LARGE \lim_{x \rightarrow 0} \frac{x}{\sqrt{1-\cos x}} \times \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}=\frac{x {\sqrt{1+\cos x }}}{\sqrt{1-\cos^2x}}\]

Answer 6

\[\LARGE \frac{x {\sqrt{1+\cos x }}}{\sqrt{\sin^2x}}= \lim_{x \rightarrow 0} \frac{x \sqrt{1+\cos x}}{\sin x}\]
and..
\[\LARGE \lim_{x \rightarrow 0} \frac{x}{\sin x}=1\]

Answer 7

\[\Huge \lim_{x \rightarrow 0} \sqrt{1+ \cos x}\]
Now you can put x=0

Answer 8

did you get anything?

Answer 9

Yes, I know what to do now, multiple by its conjugate again

Answer 10

Thanks for your help:)

Answer 11

u can just put x=0
\[\Huge \sqrt{1+\cos 0}\]
cos0=?

Answer 12

1

Answer 13

sqrt(1+1)=

Answer 14

the answer should be zero since limit x/sinx = 0

Answer 15

x/sinx=1..are you somewhat dizzy?

Answer 16

\[\lim_{x \rightarrow 0}\frac{ x }{sinx }\]=
\[\lim_{x \rightarrow 0}\frac{ 0 }{sin0 }\]=0/1=0??
If I am mistaken, please correct me

Answer 17

That is a property,you don't have to put 0 in that! You just have to remember that when x tends to 0,the expression sinx/x or x/sinx tends to 1

Answer 18

That is because for small x, sin x is approximately x, as can be seen in its power series expansion.