Question

\(\displaystyle \lim_{x\to\infty}\frac{\int_{0}^{2x} x e^{x^{2}}\,dx}{e^{4x^{2}}}\) equals

• A. \(0\)
• B. \(\infty\)
• C. \(2\)
• D. \(\dfrac{1}{2}\)

Hint:

Divide highest exponential term before taking limit.

Answer 1

The Correct Option is D

Step 1: \(\displaystyle \int x e^{x^2}dx=\frac12 e^{x^2}\)
Step 2: Numerator \[ \int_0^{2x} x e^{x^2}dx=\frac12\left(e^{4x^2}-1\right) \]
Step 3: \[ \lim_{x\to\infty}\frac{\frac12(e^{4x^2}-1)}{e^{4x^2}}=\frac12 \]