The value of `underset(n rarr oo)lim (1+2+3+…n)/(n^(2)+100)` is equal to :

Options

A. `oo`
B. `(1)/(2)`
C. 2
D. 0

Answer

To solve the limit \(\lim_{n \to \infty} \frac{1 + 2 + 3 + \ldots + n}{n^2 + 100}\), we can follow these steps:

### Step 1: Express the sum of the first \(n\) natural numbers
The sum of the first \(n\) natural numbers is given by the formula:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

### Step 2: Substitute the sum into the limit
Now, we can substitute this expression into our limit:
\[ \lim_{n \to \infty} \frac{\frac{n(n + 1)}{2}}{n^2 + 100} \]

### Step 3: Simplify the expression
We can simplify the limit:
\[ = \lim_{n \to \infty} \frac{n(n + 1)}{2(n^2 + 100)} \]
This can be rewritten as:
\[ = \lim_{n \to \infty} \frac{n^2 + n}{2(n^2 + 100)} \]

### Step 4: Factor out \(n^2\) from the numerator and denominator
Next, we factor \(n^2\) out of both the numerator and the denominator:
\[ = \lim_{n \to \infty} \frac{n^2(1 + \frac{1}{n})}{2n^2(1 + \frac{100}{n^2})} \]

### Step 5: Cancel \(n^2\) and simplify further
Now we can cancel \(n^2\) from the numerator and the denominator:
\[ = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2(1 + \frac{100}{n^2})} \]

### Step 6: Evaluate the limit as \(n\) approaches infinity
As \(n\) approaches infinity, \(\frac{1}{n}\) approaches \(0\) and \(\frac{100}{n^2}\) also approaches \(0\):
\[ = \frac{1 + 0}{2(1 + 0)} = \frac{1}{2} \]

### Conclusion
Thus, the value of the limit is:
\[ \lim_{n \to \infty} \frac{1 + 2 + 3 + \ldots + n}{n^2 + 100} = \frac{1}{2} \]