when demonstrating that $lim_{x\rightarrow2}5x + 3 = 13$ with $varepsilon = 0.3$, which of the following…

when demonstrating that $lim_{x\rightarrow2}5x + 3 = 13$ with $varepsilon = 0.3$, which of the following $delta$-values suffices? there may be more than one correct answer, so select all that apply $squaredelta = 0.02$ $squaredelta = 0.0036$ $squaredelta = 0.06$ $squaredelta = 0.12$

when demonstrating that $lim_{x\rightarrow2}5x + 3 = 13$ with $varepsilon = 0.3$, which of the following $delta$-values suffices? there may be more than one correct answer, so select all that apply $squaredelta = 0.02$ $squaredelta = 0.0036$ $squaredelta = 0.06$ $squaredelta = 0.12$

Answer

Explanation:

Step1: Recall limit - definition

The definition of $\lim_{x\rightarrow a}f(x) = L$ is: for every $\varepsilon>0$, there exists a $\delta>0$ such that if $0 < |x - a|<\delta$, then $|f(x)-L|<\varepsilon$. Here, $f(x)=5x + 3$, $a = 2$, and $L = 13$.

Step2: Calculate $|f(x)-L|$

$|f(x)-L|=|(5x + 3)-13|=|5x-10| = 5|x - 2|$.

Step3: Relate $\delta$ and $\varepsilon$

We want $|f(x)-L|<\varepsilon$, so $5|x - 2|<\varepsilon$. Given $\varepsilon=0.3$, then $|x - 2|<\frac{\varepsilon}{5}$. Substituting $\varepsilon = 0.3$, we get $|x - 2|<\frac{0.3}{5}=0.06$. So, any $\delta\leqslant0.06$ will satisfy the limit - definition.

Answer:

$\delta = 0.02$, $\delta = 0.0036$, $\delta = 0.06$