Question

Determine o limite da função f (x) = (x
2
-2x-3) / (x2
-9) para
a. x3
+
=
b. x0
+
=

Answer 1

Determine o limite da função    $f(x)=\dfrac{x^{2} -2x-3}{x^{2} -9}$   .

Questão a)

$Lim_{x\to 0} =\dfrac{x^{2} -2x-3}{x^{2} -9} ~~=~~\dfrac{0^{2} -2\cdot0-3}{0^{2} -9}~~=~~\dfrac{-3}{-9}~~=~~\dfrac{1}{3} \\ \\ \\ \\ \\ \boxed{Lim_{x\to 0} =\dfrac{x^{2} -2x-3}{x^{2} -9}~=~\dfrac{1}{3} }$

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Questão b)

$Lim_{x\to 3} =\dfrac{x^{2} -2x-3}{x^{2} -9} ~~=~~\dfrac{3^{2} -2\cdot3-3}{3^{2} -9}~~=~~\dfrac{0}{0}~~=~~indeterminacao$

Perceba que     $\dfrac{x^{2} -2x-3}{x^{2} -9}~~=~~ \dfrac{(x+1)\cdot(x-1)}{(x+3)\cdot(x-1)} ~~=~~\dfrac{(x+1)}{(x+3)}$  

Então podemos reescrever o limite da seguinte forma:

$Lim_{x\to 3}~~=~~ \dfrac{x+1}{x+3}~~=~~\dfrac{3+1}{3+3} ~~=~~\dfrac{4}{6}~~=~~\dfrac{2}{3}$

$\boxed{Lim_{x\to 3} =\dfrac{x^{2} -2x-3}{x^{2} -9}~=~\dfrac{2}{3} }$

Resposta:

a) 1/3

b) 2/3

:)