Q.If $|x - 10| < k_1 = (k_2, 10.01)$, then find the values of $k_1$ and $k_2$.

$0.3$ neighbourhood of $3$ $=\{x:|x-3|<0.3\}$.

The $\delta$-neighbourhood of a point $a$ is the open interval $(a-\delta,\,a+\delta)$, i.e. the set of all $x$ whose distance from $a$ is less than $\delta$: $|x-a|<\delta$. …

$|x-10|<\tfrac1{10}$ $=$ the $\tfrac1{10}$-neighbourhood of $10$ $=(9.9,10.1)$.

The modulus inequality $|x-a|<\delta$ is exactly the $\delta$-neighbourhood of the point $a$, namely the open interval $(a-\delta,\,a+\delta)$. Here $a=10$ and $\delta=\tfrac1{10}=0.1$, so it is the $\tfrac1{10}$-neighbourhood of $10$: …

$|x - 5| < 0.25$ is the $\delta$-neighbourhood of $5$ with radius $0.25$, written $N(5, 0.25)$ — option (d).

GSEB Class-12 Statistics, Limit chapter:

The neighbourhood notation is defined as:

$N(a, \delta) = \{x : |x - a| < \delta\}$

where $a$ is the centre and $\delta$ the radius. Comparing with: …

The $\delta$-neighbourhood of $a$ is $|x-a|<\delta$; with $a=-5,\ \delta=0.001$ this is $|x+5|<0.001$.

…

The $\delta$ neighbourhood of $a$ is $\{x : |x-a|<\delta\} = (a-\delta,\ a+\delta)$ for some $\delta>0$.

Definition. For a real number $a$ and a positive number $\delta$, the $\delta$ neighbourhood of $a$ is the set of all real numbers $x$ whose distance from $a$ is less than $\delta$:

$\{x : |x-a|<\delta\} = (a-\delta,\ a+\delta).$ …

"$x$ tends to $0$" — $x$ takes values as close to $0$ as we like, from both sides, but never equals $0$.

The symbol $x \to 0$ is read as "$x$ tends to $0$" or "$x$ approaches $0$." It means we let $x$ take a succession of values that come nearer and nearer to $0$:

• from the right ($x \to 0^{+}$): values like $0.1, 0.01, 0.001, \dots$
• from the left ($x \to 0^{-}$): values like $-0.1, -0.01, -0.001, \dots$ …