Find the limit $L$ for the given function $f$, the point $x _ { 0 }$, and the positive number $\varepsilon$. Then find a number $\delta 0$ such that, for all $x _ { t } 0

Question

Find the limit $L$ for the given function $f$, the point $x _ { 0 }$, and the positive number $\varepsilon$. Then find a number $\delta > 0$ such that, for all $x _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon$.
-f(x) = 5x + 2, L = 17, x0 = 3, and $\varepsilon$ = 0.01
A) 0.003333
B) 0.004
C) 0.002
D) 0.01

Quizplus · Accepted Answer

We have:
\begin{align*}
|f(x)-L| &= |5x+2-17|=|5x-15|=5|x-3|\
\end{align*}
So, we want to find $\delta>0$ such that $5|x-3|<\varepsilon=0.01$. 
Dividing both sides by 5, we get $|x-3|<\frac{\varepsilon}{5}=0.002$. Therefore, we can set $\delta = 0.002$. 
Thus, if $0<|x-3|<0.002$, then $|f(x)-L|<0.01$. So, the correct answer is (C).

Find the Limit LLL For the Given Function fff , the Point x0x _ { 0 }x0​

Find the Limit $L$ For the Given Function $f$ , the Point $x _ { 0 }$