Question:
The value of
\[
\lim_{n \to \infty} \frac{1 + 2 + 3 + \cdots + n}{n^2 + 100}
\]
is equal to:
The value of
\[ \lim_{n \to \infty} \frac{1 + 2 + 3 + \cdots + n}{n^2 + 100} \]
is equal to:
\[ \lim_{n \to \infty} \frac{1 + 2 + 3 + \cdots + n}{n^2 + 100} \]
is equal to:
Show Hint
Compare highest powers of n in limits at infinity.
Updated On: Mar 23, 2026
- \(\infty\)
- \(\dfrac{1}{2}\)
- \(2\)
- 0
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1:
\[ 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \]
Step 2:
\[ \frac{\frac{n(n+1)}{2}}{n^2 + 100} = \frac{n^2 + n}{2n^2 + 200} \]
Step 3: Divide numerator and denominator by \(n^2\):
\[ \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2 + \frac{200}{n^2}} = \frac{1}{2} \]
\[ 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \]
Step 2:
\[ \frac{\frac{n(n+1)}{2}}{n^2 + 100} = \frac{n^2 + n}{2n^2 + 200} \]
Step 3: Divide numerator and denominator by \(n^2\):
\[ \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2 + \frac{200}{n^2}} = \frac{1}{2} \]
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