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 $x _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon$.
-$$\mathrm { f } ( \mathrm { x } ) = \sqrt { 24 - 3 \mathrm { x } } , \mathrm { x } _ { 0 } = - 4 , \varepsilon = 0.3$$
A) $\mathrm { L } = j - 5 ; \delta = 0.57$
B) $\mathrm { L } = 6 ; \delta = 1.17$
C) $L = 7 ; \delta = 1.17$
D) $L = 6 ; \delta = 1.23$

Quizplus · Accepted Answer

First, find the limit $L$ by substituting $x_0 = -4$ into the function: $L = \sqrt{24 - 3(-4)} = \sqrt{24 + 12} = \sqrt{36} = 6$. To find $\delta$, we ensure $|f(x) - L| < \varepsilon$ leads to $|\sqrt{24 - 3x} - 6| < 0.3$, which after some algebra, gives a suitable $\delta$ close to the value in choice B.

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 $x _ { 0 }$