Question

The value of \[ \lim_{n \to \infty} \frac{3n^2 + 5n + 4}{4 + 2n^2} \] is 
 

• A. 0
• B. 0.75
• C. 1.5
• D. 3

Hint:

When finding limits, simplify the expression by dividing by the highest power of \(n\) and then apply the limit.

Answer 1

The Correct Option is C

To find the value of the limit \( \lim_{n \to \infty} \frac{3n^2 + 5n + 4}{4 + 2n^2} \), follow these steps:

1. Identify the highest power of \( n \) in the numerator and the denominator. Both the numerator \( 3n^2 + 5n + 4 \) and the denominator \( 4 + 2n^2 \) have the highest power term as \( n^2 \).

2. Divide every term in the numerator and the denominator by \( n^2 \), the highest power of \( n \) in this expression:

   \[ \frac{3n^2 + 5n + 4}{4 + 2n^2} = \frac{\frac{3n^2}{n^2} + \frac{5n}{n^2} + \frac{4}{n^2}}{\frac{4}{n^2} + \frac{2n^2}{n^2}} \]

3. Simplify the expression:

   \[ = \frac{3 + \frac{5}{n} + \frac{4}{n^2}}{\frac{4}{n^2} + 2} \]

4. As \( n \to \infty \), the terms \(\frac{5}{n}\), \(\frac{4}{n^2}\), and \(\frac{4}{n^2}\) approach zero:

   \[ = \frac{3 + 0 + 0}{0 + 2} = \frac{3}{2} \]

5. Thus, the value of the limit is \( \frac{3}{2} \), or 1.5.

Therefore, the correct answer is 1.5.