Question

Alguém poderia me ajudar a encontrar os limites dessas funções ?

Answer 1

✅ Confira abaixo as resoluções dos limites. Enviei sem querer

 

$\large\begin{array}{lr}\displaystyle\rm a)~~\lim_{x\to1} \left( x^3-3 \right) = 1^3 - 3 = -2 \\\\\red{\underline{\boxed{\boxed{\rm \therefore\:\lim_{x\to1} \left( x^3-3 \right) = -2}}}}\end{array}$

 

$\large\begin{array}{lr} \displaystyle\rm b)~~\lim_{x\to2} \sqrt{x^4 - 8} = \sqrt{2^4 - 8} = \sqrt{8} \\\\\red{\underline{\boxed{\boxed{\rm \therefore\:\lim_{x\to2} \sqrt{x^4 - 8} = \sqrt{8}}}}}\end{array}$

 

$\large\begin{array}{lr}\begin{aligned}\displaystyle\rm c)~~\lim_{x\to2}\sqrt{\frac{x^3+2x+3}{x^2+5}} &=\\\\ &=\displaystyle\rm\sqrt{\lim_{x\to2} \frac{x^3+2x+3}{x^2+5} } =\\\\ &=\displaystyle\rm\sqrt{ \frac{\lim_{x\to2}\,x^3+2x+3}{\lim_{x\to2}\,x^2+5}} = \\\\ &= \sqrt{\dfrac{2^3+2\cdot 2+3}{2^2 + 5}} \end{aligned}\\\\\large\red{\underline{\boxed{\boxed{\rm \therefore\:\lim_{x\to2}\sqrt{\frac{x^3+2x+3}{x^2+5}} = \sqrt{\dfrac{15}{9} } = \dfrac{\sqrt{15}}{3} }}}}\end{array}$

 

$\large\begin{array}{lr}\displaystyle\rm d)~~ \lim_{x\to -3} \frac{x^2 - 9}{x+3} = \frac{0}{0}~~\boxtimes  \\\\ \begin{aligned} \displaystyle\rm \lim_{x\to -3} \frac{x^2 - 9}{x+3} &= \displaystyle\rm \lim_{x\to -3} \frac{x^2 - 3^2}{x+3}\\\\ &= \displaystyle\rm \lim_{x\to -3} \frac{(x-3)\cancel{(x+3)}}{\cancel{(x+3)}} \\\\ &=\displaystyle\rm\lim_{x\to-3} \,( x-3 ) \\\\ &= \displaystyle\rm -3-3 \end{aligned} \\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \lim_{x\to -3} \frac{x^2 - 9}{x+3} = -6 }}}}\end{array}$

 

$\large\begin{array}{lr}\displaystyle\rm e) \lim_{x\to ^1\!/_3} \frac{3x^2 -x}{3x - 1} = \frac{0}{0}~~\boxtimes \\\\\begin{aligned}\displaystyle\rm \lim_{x\to ^1\!/_3} \frac{3x^2 -x}{3x - 1} &=\displaystyle\rm \lim_{x\to ^1\!/_3} \frac{3x^2 -x}{3x - 1} \\\\&=\displaystyle\rm \lim_{x\to ^1\!/_3} \frac{x\cancel{(3x -1)}}{\cancel{(3x - 1)}} \\\\&=\displaystyle\rm \lim_{x\to ^1\!/_3} x  \end{aligned}\\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \lim_{x\to ^1\!/_3} \frac{3x^2 -x}{3x - 1} = \dfrac{1}{3} }}}} \end{array}$

 

$\large\begin{array}{lr}\displaystyle\rm f)~~\lim_{x\to 3} \frac{x^3 -27}{x-3} = \frac{0}{0}~~\boxtimes\\\\\begin{aligned}\displaystyle\rm \lim_{x\to 3} \frac{x^3 -27}{x-3}&=\displaystyle\rm \lim_{x\to3} \frac{x^3 - 3^3}{x-3} \\\\&=\displaystyle\rm \lim_{x\to3} \frac{\cancel{(x - 3)}\left( x^2 + 3x + 3^2 \right)}{ \cancel{(x-3)}} \\\\&=\displaystyle\rm \lim_{x\to3}\,\left( x^2 + 3x + 9 \right) \\\\&=\displaystyle\rm 3^2 + 3\cdot 3 + 9 \end{aligned} \\\\\red{\underline{\boxed{\boxed{\rm \therefore\: \lim_{x\to 3} \frac{x^3 -27}{x-3} = 27 }}}} \end{array}$