Consider $\lim _ { x \rightarrow - 3 } ( | x | + 1 ) = 4$ In order for $| ( | x | + 1 ) - 4 |

Question

Consider $\lim _ { x \rightarrow - 3 } ( | x | + 1 ) = 4$ In order for $| ( | x | + 1 ) - 4 | < 0.01$ whenever $0 < | x - ( - 3 ) | < \delta$ which of the following is true for the largest ?
A)0.001
B)0.003
C)0.01
D)0.03
E)Impossible to find

Quizplus · Accepted Answer

Given the limit condition, we need to find a $\delta$ such that $| ( | x | + 1 ) - 4 | < 0.01$ whenever $0 < | x + 3 | < \delta$. This simplifies to $| | x | - 3 | < 0.01$. Since $x$ is approaching -3, we consider the negative side, where $| x | = -x$ for $x < 0$. Thus, we have $| -x - 3 | < 0.01$, which simplifies to $| -(x + 3) | < 0.01$. This means $x + 3$ must be within 0.01 of 0, so $\delta = 0.01$ is the largest value that satisfies the condition.

Consider lim⁡x→−3(∣x∣+1)=4\lim _ { x \rightarrow - 3 } ( | x | + 1 ) = 4limx→−3​(∣x∣+1)=4

Consider $\lim _ { x \rightarrow - 3 } ( | x | + 1 ) = 4$