Question

Mostre que lim (5x -3) = 2
x -> 1

Answer 1

Olá!

Definição de limite: $\displaystyle \lim_{x \rightarrow a} f(x) = L \Longleftrightarrow (\forall \epsilon >0{, } \exists \delta >0 \mid 0<|x -a| < \delta \implies |f(x)- L|<\epsilon ).$

Devemos mostrar que, para qualquer $\epsilon >0$, existe $\delta > 0$ tal que

$0< |x - 1| < \delta \implies |(5x-3)-2|< \epsilon$

Note que

$|(5x-3)-2| < \epsilon \Longleftrightarrow |5x-5|< \epsilon \Longleftrightarrow \\ \Longleftrightarrow |5(x-1)|< \epsilon \Longleftrightarrow |x-1|< \dfrac{\epsilon}{5}$

Assim, se escolhermos $\delta = \dfrac{\epsilon}{5}$, teremos:

$\forall \epsilon > 0{, } \exists \delta = \dfrac{\epsilon}{5} > 0 \mid 0 <|x -1|<\delta \implies |(5x-3) -2|< \epsilon$

De fato,

$0<|x-1| < \dfrac{\epsilon}{5}| \implies \\ \implies |x-1|<\dfrac{\epsilon}{5} \implies 5|x-1|<\epsilon \implies \\ \implies |5x-5| < \epsilon \implies |(5x-3)-2|< \epsilon \blacksquare$

Logo,

$\displaystyle \lim_{x \rightarrow 1}(5x-3) = 2.$