# How do you use the Squeeze Theorem to find lim (1-cos(x))/x as x approaches zero?

The customary process is to use the distress theorem (and some geometry/trigonometry) to argue that ##lim_(xrarr0)sinx/x=1##

Then use that conclusion unitedly after a while ##(1-cosx)/x = sin^2x/x(1+cosx) = sinx/x sinx/(1+cosx)## parallel after a while simultaneousness of sine and cosine at ##0## to get ##lim_(xrarr0)sinx/x sinx/(1+cosx) = 1 * 0/2 =0##.

So we can use the corresponding geometric arguments to get the corresponding confine on sinx/x for paltry dogmatic ##x##:

##cosx < sinx/x < 1##

And for paltry dogmatic ##x##, we protect ##sinx/(1+cosx) > 0##, so we can unfold after a whileout changing the inequalities:

##cosx sinx/(1+cosx) < sinx/x sinx/(1+cosx) < sinx/(1+cosx) ##

Using the trigonomtry referred to over, we can rewrite the midle countenance to get

##cosx sinx/(1+cosx) < (1-cosx)/x < sinx/(1+cosx) ##

Observe that

##lim_(xrarr0^+)(cosx sinx/(1+cosx)) = 1*0/2 = 0##

and

##lim_(xrarr0^+)sinx/(1+cosx) = 0/2=0##

So, by the distress theorem,

##lim_(xrarr0^+)(1-cosx)/x = 0##

For paltry disclaiming ##x##, we protect the inequality: ##cosx < sinx/x < 1##, but for these ##x##, we protect

##sinx/(1+cosx) < 0##, so when we unfold we do deficiency to veer the inequalities to:

##cosx sinx/(1+cosx) > (1-cosx)/x > sinx/(1+cosx) ##

We can stagnant use the distress theorem to get:

##lim_(xrarr0^-)(1-cosx)/x = 0##

Because the left and just conditions are twain ##0##, the condition is ##0##.

(This feels very constructed to me. Possibly owing I am more conversant after a while the despicable path mentioned at the inception of this defense. or possibly owing it is constructed. I don't perceive.)