Answer to Question #259010 in Calculus for Pankaj

Question

Using the Epsilon and Delta definition, show that lim x approaching 2 for (3x-5) is =1

Answer 1

$Given \ any \ \epsilon \gt0 \ we \ choose\ \delta_\epsilon \lt \frac{\epsilon}{3}$

Explanation

we have to show or prove that given any $\epsilon \gt0$ we can find a $\delta_\epsilon \gt 0$ such that:

$\lvert f(x) -1 \rvert \lt \epsilon \space \ for \ x \isin (2- \delta_\epsilon , 2+ \delta_\epsilon )$

we evaluate the difference:

$\lvert f(x) -1 \rvert = |3x-5-1|=|3x-6|\\=3|x-2|$

and we can see that :

$|f(x)-1|\lt \epsilon \iff |x-2| \lt \frac{\epsilon}{3}$

so given $\epsilon \gt 0$ we can choose $\delta_\epsilon \lt \frac{\epsilon}{3}$ and we have:

$x \isin (2- \delta_\epsilon , 2+ \delta_\epsilon ) \implies |x-2| \lt \frac{\epsilon}{3}\\\iff |f(x)-1|\lt\epsilon$

which shows the point.