Question

Use the precise definition of a limit to prove that lim x->2 5x-7=3, pic also included

Answer 1

We need to use the formal definition of a limit, The definition is:

Let be L be a real number. Then:

$\lim_{x\to a}f(x)=L$

if, for every ε > 0, there exists a δ > 0 such that if 0 < |x - a| < δ, then |f(x) - L| < ε

In this case, we need to prove:

$\lim_{x\to2}(5x-7)=3$

Then, we need to find a relationship between ε and δ, so for any value of ε> 0 we can find a value of δ.

We know that, for a given ε > 0, |(5x -7) - 3| < ε, and 0 < |x - 2| < δ

Then:

$\begin{gathered} |5x-10|<\varepsilon \\ 5|x-2|<\varepsilon \end{gathered}$

Now we can divide each side by 5.

$|x-2|<\frac{\varepsilon}{5}$

We are now really close from relate the ε and δ.

We can use the triangle inequality:

$|a+b|<|a|+|b|$

Then, we can add and subtract 1 inside the absolute value:

$\lvert x-2-1+1\rvert\lt\frac{\varepsilon}{5}$

We have:

$\lvert x-3+1\rvert\lt\frac{\varepsilon}{5}$

By the triangle inequality:

$|x-3|+|1|<\lvert x-2\rvert\lt\frac{\varepsilon}{5}$

And since the absolute value of any number is bigger or equal than 0:

$|x-3|<|x-3\rvert+\lvert1\rvert\lt\lvert x-2\rvert\lt\frac{\varepsilon}{5}$

And we have:

$|x-3|<\frac{\varepsilon}{5}$

And since ε > 0, ε/5 > 0, we can define:

$|x-3|<\delta=\frac{\varepsilon}{5}$

And we have proven that:

For a given ε > 0, exists δ = ε/5 such that, if 0 < |x - 2| < δ = ε/5, then |f(x) - L| < ε