Matematické Fórum

Pavel · 03. 12. 2014 01:02 — Editoval Pavel (03. 12. 2014 01:03)

Find the limit

$\lim_{x\to 0}\frac{\sin x-x}{x^3}.$

Only the identity

$\lim_{x\to 0}\frac{\sin x}x=1$

and elementary manipulations are allowed. Advanced techniques based on derivatives are not permitted.

jarrro · 03. 12. 2014 14:12 — Editoval jarrro (27. 02. 2023 21:06)

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Skrytý text:

\begin{array}{r} l = lim_{x \to 0} \frac{\sin (x) - x}{x^{3}} = \\ = 4 \cdot lim_{x \to 0} \frac{\sin (2 x) - 2 x + 2 x - 2 x \cos (x)}{8 x^{3}} = \\ = 4 (l + \frac{1}{4} lim_{x \to 0} \frac{1 - \cos (x)}{x^{2}}) \\ l = 4 l + \frac{1}{2} \Rightarrow l = - \frac{1}{6} \end{array}

Pavel · 03. 12. 2014 16:03 — Editoval Pavel (03. 12. 2014 16:05)

↑ jarrro:

Correct solution. I used a similar idea:

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jarrro · 04. 12. 2014 09:46

↑ Pavel:Nice. Your solution is more elegant because it is builded on equality of whole functions not only limit equality

vanok · 04. 12. 2014 23:20 — Editoval vanok (05. 12. 2014 13:51)

Hi Pavel
Annoter variant.

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Cordially

Marian · 05. 12. 2014 08:15

How do you know that the given limit:

(i) exists,
(ii) is finite? (It is clear that the constructed linear equations (all three) have more than the only solution.)

jarrro · 05. 12. 2014 09:18 — Editoval jarrro (05. 12. 2014 09:19)

↑ Marian:function is even, negative, bounded and continous on $$-\infty,0$\cup$0,\infty$$ . Moreover it is on $$-\infty, 0$$ dereasing and on $$0,\infty$$ is increasing.
but I cannot prove this without differential calculus

Pavel · 05. 12. 2014 10:40

↑ Marian:

The existence of the limit was assumed implicitly. Although I agree that your questions are important.

Marian · 05. 12. 2014 10:40

↑ jarrro:

Exactly, this is the hidden problem.

Matematické Fórum

#1 03. 12. 2014 01:02 — Editoval Pavel (03. 12. 2014 01:03)

Limit - elementary solution

#2 03. 12. 2014 14:12 — Editoval jarrro (27. 02. 2023 21:06)

Re: Limit - elementary solution

#3 03. 12. 2014 16:03 — Editoval Pavel (03. 12. 2014 16:05)

Re: Limit - elementary solution

#4 04. 12. 2014 09:46

Re: Limit - elementary solution

#5 04. 12. 2014 23:20 — Editoval vanok (05. 12. 2014 13:51)

Re: Limit - elementary solution

#6 05. 12. 2014 08:15

Re: Limit - elementary solution

#7 05. 12. 2014 09:18 — Editoval jarrro (05. 12. 2014 09:19)

Re: Limit - elementary solution

#8 05. 12. 2014 10:40

Re: Limit - elementary solution

#9 05. 12. 2014 10:40

Re: Limit - elementary solution

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