Question

Determine os seguintes limites (não pode usar a Regra de L'Hospital)

$\lim_{x \to \ 0 } \frac{2x^2-x-3}{3x^2+8x+5}$


$\lim_{x \to \ 0 } \frac{x}{sen 3x}$


$\lim_{x \to +\infty} (\sqrt{x+1}- \sqrt{x} )$

Answer 1

a)

Lim (2x²-x-3)/(3x²+8x+5)

x-->0

ax²+bx+c=a*(x-x')*(x-x'')   a ≠0 e x' e x'' são raízes

2x²-x-3=2*(x+1)(x-1/3)

3x²+8x+5=3*(x+1)*(x+5/3)

Lim 2(x+1)(x-3/2)/3*(x+1)*(x+5/3)

x-->0

Lim 2(x-1/3)/3*(x+5/3)

x-->0

2(0-3/2)/3*(0+5/3)

(-3)/5 =-3/5  é a resposta

b)

Lim x/(sen(3x))

x-->0

Lim(3x)/(3x) * x/(sen(3x))

x-->0

Lim(x)/(3x) * 3x/(sen(3x))

x-->0

***********************

Limite fundamental

Lim  3x/(sen(3x) =1

x-->0

***********************

Lim(x)/(3x)  

x-->0

Lim(1)/(3)   = 1/3  é a resposta

x-->0

c)

Lim (√(x+1) -√x)

x-->∞

Lim (√(x+1) -√x)*(√(x+1) +√x)/(√(x+1) +√x)

x-->∞

Lim (√(x+1)² -√x²)/(√(x+1) +√x)

x-->∞

Lim (x+1 -x)/(√(x+1) +√x)

x-->∞

Lim  1/(√(x+1) +√x)

x-->∞

= 1/(√(∞+1) +√∞) =1/∞ = zero é a resposta  

Answer 2

$\mathsf{\lim_{x \to 0}\dfrac{2{x}^{2}-x-3}{3{x}^{2}+8x+5}=\dfrac{2.{0}^{2}-0-3}{3.{0}^{2}+8.0+5}}\\\huge\mathsf{\lim_{x \to 0}\dfrac{2{x}^{2}-x-3}{3{x}^{2}+8x+5}=-\dfrac{3}{5}}$

$\mathsf{\lim_{x to 0}\dfrac{x}{\sin(3x)}\\\mathsf{\lim_{x \to 0}\dfrac{\frac{x}{x}}{\frac{\sin(3x)}{3x}}}=\dfrac{1}{3}}$

$\mathsf{\lim_{x \to +\infty}(\sqrt{x+1}-\sqrt{x})=\lim_{x \to +\infty}\dfrac{x+1-x}{\sqrt{x+1}+\sqrt{x}}=\dfrac{1}{\infty}=0}$