Limits, Continuity & Differentiability
General
Grade 12
Question:
<p>For each t
∈
R,
let [t] be the greatest integer less than or equal to t.
Then
limx→0+ x
1
x
+
2
x
+ · · · +
15
x
:
(1) is equal to 15.
(2) is equal to 120.
(3) does not exist.
(4) is equal to 0.</p>
is equal to 15
is equal to 120
does not exist
is equal to 0
Step-by-Step Solution
Key Concept: Use the Squeeze Theorem property of GIF: t -1 < [t] \leqt.
<p><strong>1</strong>: We know k</p> x -1 < k x \leqk x.<p><strong>2</strong>: Summing for k = 1 to 15: P k</p> x -15 < P k x \leqP k x.<p><strong>3</strong>: Multiply by x (where x > 0): P k -15x < x P k</p> x \leqP k.<p><strong>4</strong>: As x \to 0+, both sides approach P15</p> k=1 k = 15 \times 16 2 = 120.
Correct Answer: (2)