Prima limită:
[tex]\lim\limits_{x\to 0}\Big(\dfrac{\tan x}{x}\Big)^{\dfrac{1}{\sin^2 x}} = \\ \\ \dfrac{1}{\sin^2 x} = \dfrac{x^2}{x^2 \cdot \sin^2 x}=\dfrac{1}{x^2}\cdot \Big(\dfrac{x}{\sin x}\Big)^2\to\dfrac{1}{x^2}\\ \\=\lim\limits_{x\to 0}\Big(1+\dfrac{\tan x-x}{x}\Big)^{\dfrac{1}{x^2}}= \\ \\ =\lim\limits_{x\to0}\Big(1+\dfrac{\tan x-x}{x}\Big)^{\dfrac{x}{\tan x- x}\cdot \dfrac{\tan x- x}{x}\cdot \dfrac{1}{x^2}}=[/tex]
[tex]=e^{\Big{\lim\limits_{x\to 0}}\dfrac{\tan x - x}{x^3}} =e^{\Big{\lim\limits_{x\to 0}}\dfrac{1+\tan^2 x - 1}{3x^2}}= e^{\Big{\lim\limits_{x\to 0}}\dfrac{\tan^2 x}{3x^2}}= \\ \\ =e^{\Big{\lim\limits_{x\to 0}}\dfrac{1}{3}\cdot \Big(\dfrac{\tan x}{x}\Big)^2} = e^{\dfrac{1}{3}} = \sqrt[3]{e}[/tex]
A doua limită:
[tex]l = \lim\limits_{x\to \infty} \left(x-\sqrt{x^2+x+1}\,\dfrac{\ln\left(e^x+x\right)}{x}\right)\\ \\ x\to \infty: \\ \\\Rightarrow e^x+x\approx e^x\Rightarrow \ln(e^x+x)\approx \ln(e^x)=x\\\Rightarrow \sqrt{x^2+x+1}\approx x+\dfrac{1}{2}\\ \text{Deoarece }x+\dfrac{1}{2}\text{ e asimptota oblica la }+\infty \text{ pentru } \sqrt{x^2+x+1}[/tex]
[tex]\Rightarrow l = \lim\limits_{x\to \infty}\left[x-\Big(x+\dfrac{1}{2}\Big)\cdot \dfrac{x}{x}\right] = \lim\limits_{x\to \infty}\left[x-\Big(x+\dfrac{1}{2}\Big)\right] = \boxed{-\dfrac{1}{2}}[/tex]
