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Sa se calculeze \( \displaystyle \lim_{x\mapsto 0} \frac{tgx -x}{x^{2}} \)

Andrei Ciupan wrote:Sa se calculeze \( \displaystyle \lim_{x\mapsto 0} \frac{tgx -x}{x^{2}} \)

La Multi Ani, Andrei ! Ma bucur de frumoasele tale rezultate in competitiile scolare si ca participi la acest forum prin propunerea/rezolvarea unor probleme interesante, de calitate. Mult succes in continuare !
Sa revenim la problema propusa care a fost data cu multi ani in urma la olimpiada cu mentiunea expresa de a nu se folosi regulile lui l'Hospital.

Notam \( f(x)=\frac{tgx -x}{x^{2}}\ ,\ x\in\left(-\epsilon , \epsilon ) \) unde \( \epsilon\in\left(0\ ,\ \frac {\pi}{2}\right)\ . \) Se stie ca \( \lim_{x\to 0}\ \frac {\tan x}{x}=1\ , \)

\( \lim_{x\to 0}\ \frac {1-\cos x}{x^2}=\frac 12 \) si \( \underline {\overline {\left\|\ x\in\left(0\ ,\ \frac {\pi}{2}\right)\ \Longrightarrow\ 0\ <\ \sin x\ <\ x\ <\ \tan x\ \right\|}}\ . \) Se arata usor ca

\( 0\ <\ \frac {\tan x-x}{x^2}\ <\ x\cdot\frac {\tan x}{x}\cdot\frac {1-\cos x}{x^2}\ . \) Asadar, \( \lim_{x\searrow 0}\ \frac {\tan x-x}{x^2}=0\ \Longrightarrow\ f(0+0)=0\ . \)

Insa \( \lim_{x\nearrow 0}\ \frac {\tan x-x}{x^2}\ \stackrel{(x:=-x)}{\ =\ }\ -\lim_{x\searrow 0}\ \frac {\tan x-x}{x^2}=0\ \Longrightarrow\ f(0-0)=0\ . \) In concluzie,

\( f(0-0)\ =\ f(0+0)\ =\ 0\ \Longleftrightarrow\ \lim_{x\to 0}\ f(x)\ =\ 0\ , \) adica \( \lim_{x\to 0}\ \frac {\tan x-x}{x^2}\ =\ 0\ . \)