Salut,
[tex]\displaystyle\lim\limits_{x\to 2}\dfrac{x^5-32}{x^4-16}=\lim\limits_{x\to 2}\dfrac{(x-2)(x^4+2x^3+4x^2+8x+16)}{(x-2)(x^3+2x^2+4x+8)}=\\\\\\=\lim\limits_{x\to 2}\dfrac{x^4+2x^3+4x^2+8x+16}{x^3+2x^2+4x+8}=\\\\\\=\dfrac{2^4+2\cdot 2^3+4\cdot 2^2+8\cdot 2+16}{2^3+2\cdot 2^2+4\cdot 2+8}=\dfrac{5\cdot 2^4}{4\cdot 2^3}=\dfrac{5}2.[/tex]
Am folosit formula de calcul prescurtat:
aⁿ -- bⁿ = (a -- b)(aⁿ⁻¹ + aⁿ⁻²b + aⁿ⁻³b² + ... + abⁿ⁻² + bⁿ⁻¹).
Green eyes.
