To solve the problem [tex]\( 5 \lim _{x \rightarrow 9^{+}}\lfloor x\rfloor - 3 \lim _{x \rightarrow 3^{-}}\lfloor x\rfloor \)[/tex], we will methodically evaluate each limit and then simplify the expression using these values.
Recall that the floor function, denoted as [tex]\(\lfloor x \rfloor\)[/tex], represents the greatest integer that is less than or equal to [tex]\(x\)[/tex].
1. Calculate [tex]\(\lim_{x \to 9^+} \lfloor x \rfloor\)[/tex]:
- When [tex]\(x\)[/tex] approaches 9 from the right ([tex]\(x \to 9^+\)[/tex]), [tex]\(x\)[/tex] is slightly greater than 9.
- The floor function [tex]\(\lfloor x \rfloor\)[/tex] will yield the greatest integer less than or equal to [tex]\(x\)[/tex].
- For [tex]\(x\)[/tex] just greater than 9 (e.g., 9.001), [tex]\(\lfloor x \rfloor\)[/tex] is 9.
Therefore, [tex]\(\lim_{x \to 9^+} \lfloor x \rfloor = 9\)[/tex].
2. Calculate [tex]\(\lim_{x \to 3^-} \lfloor x \rfloor\)[/tex]:
- When [tex]\(x\)[/tex] approaches 3 from the left ([tex]\(x \to 3^-\)[/tex]), [tex]\(x\)[/tex] is slightly less than 3.
- The floor function [tex]\(\lfloor x \rfloor\)[/tex] will yield the greatest integer less than or equal to [tex]\(x\)[/tex].
- For [tex]\(x\)[/tex] just less than 3 (e.g., 2.999), [tex]\(\lfloor x \rfloor\)[/tex] is 2.
Therefore, [tex]\(\lim_{x \to 3^-} \lfloor x \rfloor = 2\)[/tex].
3. Substitute these limit values into the expression:
- The expression is [tex]\(5 \lim_{x \to 9^+} \lfloor x \rfloor - 3 \lim_{x \to 3^-} \lfloor x \rfloor\)[/tex].
- Substituting the calculated limits, we get [tex]\(5 \cdot 9 - 3 \cdot 2\)[/tex].
4. Calculate the final value:
- [tex]\(5 \cdot 9 = 45\)[/tex],
- [tex]\(3 \cdot 2 = 6\)[/tex],
- Therefore, [tex]\(45 - 6 = 39\)[/tex].
Hence, the value of [tex]\(5 \lim_{x \to 9^+} \lfloor x \rfloor - 3 \lim_{x \to 3^-} \lfloor x \rfloor\)[/tex] is [tex]\(\boxed{39}\)[/tex].
