# Calc 101 Limit Question

Lim 1 - cos(x)/x^2
x->0

My attempt:

lim [1 - cos (x)/ x][lim 1/x] = (1)(1/0)

lim [1 - cos (x)/ x]=1?
How ??
While solving limit Problems, you make use of certain solutions
like
$$\lim_{x\rightarrow 0} \frac{Sin(x)}{x}=1$$

$$\lim_{x\rightarrow 0} \frac{e^x-1}{x}=1$$

==============================================
eg:$$\lim_{x\rightarrow 0} \frac{Tan(x)}{x}$$

=>$$\lim_{x\rightarrow 0} \frac{\frac{Sin(x)}{cosx}}{x}$$

==>$$\lim_{x\rightarrow 0} \frac{Sin(x)}{x} \lim_{x\rightarrow 0} \frac{1}{Cosx}$$
==>1*1=1
end example==================
In this case,try to convert this into a known limit
HINT:$$cos2x=1-2sin^2x$$

Last edited:
use le'hopitals(spelling?) rule

Last edited:
berkeman
Mentor
poolwin2001 said:
lim [1 - cos (x)/ x]=1?

Hmm, I got 1/2. I think you left the x^2 out of the original denominator:

lim x-->0 of ( 1-cos(x) )/x^2

= lim x-->0 ( sin(x) )/2x

= lim x-->0 ( cos(x) )/2 = 1/2

BTW, here's a good page on L'Hopital's Rule:

http://www.math.hmc.edu/calculus/tutorials/lhopital/

PS -- I went back and used a calculator to plug in small numbers for x, and it looks more like the limit of the original function of x goes to zero. I wonder if I did the first differentiation wrong, or the calculator is fooling me. What do other folks get?

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I guess this can be solved in eitherof following 2 ways.:

1. Using formula of cosx=1-(x^2)/2!+(x^4)/4!...
so that
(1-cosx)=x^/2-(x^4)/4!....
=x^2[1-(x^2)/4!....]
so that
Lt x->0 of (1-cosx)/x^2 becomes :
=Lt x->0 of 1/2[1-(x^2)/4!....]
=1/2

2. other method could be usin L'Hospital's rule using differentiation
this too yields answer as 1/2

take care

PS -- I went back and used a calculator to plug in small numbers for x, and it looks more like the limit of the original function of x goes to zero. I wonder if I did the first differentiation wrong, or the calculator is fooling me. What do other folks get?