limits: prove lim[x->0] (1 - cosx)/X = 0

[sungjin6458]

Prove lim(x->0) (1-cosx)/X = 0 using standard trig

I got:

lim [(1-cosx)/x][(1+cosX)/(1+cosx)]

lim [(1-cos^2x0)/(x(1+cosx))]

lim [sin^2x/(x(1+cosx))]

I don't know what to do next.
:roll: thanks for the help in advance...

 

[tkhunny]

Are you waiting for bells to start ringing. An important part of a "proving" problem is knowing when you are done. Why do you think you are not done? Everything in the Numerator is finite and well-behaved.

 

[stapel]

Assuming you mean "x" and "X" to be the same variable, you have:

. . . . .[1 - cos(x)] / x

You've manipulated this to get:

. . . . .[sin²(x)] / [x (1 + cos(x))]

Split this up:

. . . . .[sin(x) / x] [sin(x) / (1 + cos(x))]

What is the limit for "sin(x) / x"? (You can just evaluate the other bit, since there is no division-by-zero problem.)

Eliz.

 

[sungjin6458]

sorry about taht...

yeah sorry i didnt clarify, but altho i know taht (sinx)/x =1
my teacher wants to me to prove this using another way.

 

[galactus]

You have an indeterminate form. Are you allowed to use L'Hopitals Rule?.


\(\displaystyle \L\\\frac{d}{dx}[1-cos(x)]=sin(x)\)

Of course, the derivative of x is 1.

You have:

\(\displaystyle \L\\\lim_{x\rightarrow{0}}(sin(x))\)

You can see what this limit is, can't you?.