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) = -10x + 5, L = -15, x0 = 2, and $\varepsilon$ = 0.01
A) 0.004
B) 0.001
C) 0.002
D) -0.005

Quizplus · Accepted Answer

We know that the limit of $f$ as $x$ approaches $x_0$ is $L$, which means that as $x$ gets closer and closer to $x_0$, the value of $f(x)$ gets closer and closer to $L$. In other words, for any positive value of $\varepsilon$, we can find a positive value of $\delta$ such that when $x$ is within $\delta$ of $x_0$, then $f(x)$ is within $\varepsilon$ of $L$.

So in this case, we want to find a positive value of $\delta$ such that when $x$ is within $\delta$ of $2$, then $f(x)$ is within $0.01$ of $-15$. Let's start by setting up the inequality:

$$|f(x) - L| < \varepsilon$$
$$|-10x + 5 - (-15)| < 0.01$$
$$|-10x + 20| < 0.01$$
$$-0.01 < -10x + 20 < 0.01$$
$$19.99 < 10x < 20.01$$
$$1.999 < x < 2.001$$

So we know that whenever $x$ is within $0.001$ of $2$, the value of $f(x)$ will be within $0.01$ of $-15$. Therefore, we can choose $\delta = 0.001$.

Note that choice B is the only value close to the exact value of $\delta$ we found.

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 $x _ { 0 }$