Question

Resolve este Limite passo-a-passo:

Answer 1

$\mathrm{\lim_{x\to0}\dfrac{x(x^2-1)}{x^2}=\lim_{x\to0}\dfrac{x^2-1}{x}=\lim_{x\to0}\bigg(x-\dfrac{1}{x}\bigg)}\\\\ \mathbf{Com\ x\to0\ pela\ direita:}\\\\ \mathrm{\lim_{x\to0+}\bigg(x-\dfrac{1}{x}\bigg)=\lim_{x\to0+}(x)-\lim_{x\to0+}\bigg(\dfrac{1}{x}\bigg)=0-\infty=\boxed{\mathrm{-\infty}}}\\\\ \mathbf{Com\ x\to0\ pela\ esquerda:}\\\\ \mathrm{\lim_{x\to0-}\bigg(x-\dfrac{1}{x}\bigg)=\lim_{x\to0-}(x)-\lim_{x\to0-}\bigg(\dfrac{1}{x}\bigg)=0-(-\infty)=\boxed{\mathrm{+\infty}}}$

$\mathbf{Portanto:}\\\\ \mathrm{J\acute{a}\ que\ \lim_{x\to0+}f(x)\neq\lim_{x\to0-}f(x)\ \to\ \boxed{\boxed{\mathbf{\nexists\lim_{x\to0}f(x)}}}}$

Answer 2

Lim x(x²-1)/x²

x→0

Lim (x²-1)/x = -1/0 = -∞

x→0