We have an answer to "How can we verify this limit via ϵ−δ..."

zagonek34

Answered question

2022-01-12

How can we verify this limit via

ϵ - δ

method?
I have to calculate the limit

lim_{x \to 0} \frac{1 - \cos (x)}{3 \sin^{2} (x)}

Answer & Explanation

Lakisha Archer

Beginner2022-01-13Added 39 answers

We have to show that

\forall ϵ > 0 : \exists δ \in (0, \frac{π}{2}] : \forall x \in R {0}

: (| x | < δ \Rightarrow | f (x) | < ϵ)

where

f (x) = \frac{{(1 - \cos x)}^{2}}{6 \sin^{2} x} = \frac{{(1 - \cos x)}^{2}}{6 (1 - \cos^{2} x)}

Let

t = \cos x

. Then, because

t \neq 1 whenever x \neq 0 and | x | < \frac{π}{2}

f (x) = \frac{{(1 - t)}^{2}}{6 (1 - t^{2})} = \frac{(1 - t)}{6 (1 + t)} = \frac{(2 - 1 - t)}{6 (1 + t)}

= \frac{1}{3 (1 + t)} - \frac{1}{6}

f(x) strictly decreases w.r.t.t.
Find t that makes

f (x) = ϵ

. After all,

t = \frac{2}{(6 ϵ + 1)} - 1

. Since

t = \cos x

, choosing

δ = \arccos (\frac{2}{6 min {ϵ, \frac{1}{6}} + 1} - 1)

proves the claim. I think you can show that

δ

is an increasing function of

ϵ

Mary Nicholson

Beginner2022-01-14Added 38 answers

You can also multiply and divide by

1 + \cos x

lim_{x \to 0} \frac{1 - \cos x}{3 \sin^{2} x} = lim_{x \to 0} \frac{1 - \cos^{2} x}{3 \sin^{2} x (1 + \cos x)} =

lim_{x \to 0} \frac{\sin^{2} x}{3 \sin^{2} x (1 + \cos x)} = \frac{1}{3} lim_{x \to 0} \frac{1}{1 + \cos x} = \frac{1}{6}

If you need to use the definition, you can easily see that

| \frac{1 - \cos x}{3 \sin^{2} x} - \frac{1}{6} | = \frac{1}{3} | \frac{1}{1 + \cos x} - \frac{1}{2} | \to 0 (x \to 0)

Since there is no indetermination in this limit, you can use Heine's definition in a very straightforward way (instead of Cauchy's). Or, you can go on to obtain

| \frac{1}{1 + \cos x} - \frac{1}{2} | = \frac{1}{2} | \frac{1 - \cos x}{1 + \cos x} | \leq \frac{1}{2} | 1 - \cos x | \leq \frac{1}{2} | x |

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