Question

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

Answer 1

The usual procedure is to use the squeeze theorem (and some geometry/trigonometry) to prove that `lim_(xrarr0)sinx/x=1`

Then use that result together with `(1-cosx)/x = sin^2x/x(1+cosx) = sinx/x sinx/(1+cosx)` along with continuity of sine and cosine at `0` to get
`lim_(xrarr0)sinx/x sinx/(1+cosx) = 1 * 0/2 =0`.

So we can use the same geometric arguments to get the same bounds on sinx/x for small positive `x`:

`cosx < sinx/x < 1`

And for small positive `x`, we have `sinx/(1+cosx) > 0`, so we can multiply without changing the inequalities:

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

Using the trigonomtry referred to above, we can rewrite the midle expression 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 squeeze theorem,

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

For small negative `x`, we keep the inequality: `cosx < sinx/x < 1`, but for these `x`, we have

`sinx/(1+cosx) < 0`, so when we multiply we do need to change the inequalities to:

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

We can still use the squeeze theorem to get:

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

Because the left and right limits are both `0`, the limit is `0`.

(This feels very artificial to me. Perhaps because I am more familiar with the common approach mentioned at the beginning of this answer. or perhaps because it is artificial. I don't know.)