Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be a differentiable function with \( \lim_{x \to \infty} f(x) = \infty \) and \(\lim_{x \to \infty} f'(x) = 2. \) Then \[ \lim_{x \to \infty} \left( 1 + \frac{f(x)}{x^2} \right)^x \] equals ..............

Hint:

When dealing with limits involving functions whose derivatives approach a constant, approximate the function and analyze the behavior of the resulting expression.

Answer 1

Correct Answer: 7.25 - 7.5

Step 1: Asymptotic form of \(f(x)\)

Since \(f'(x)\to 2\), by standard results on asymptotic behavior,
\[f(x)=2x+o(x)\quad \text{as } x\to\infty.\]

Hence,
\[\frac{f(x)}{x^2} \frac{2}{x}+o!\left(\frac{1}{x}\right).\]

Step 2: Evaluate the limit

\[\left(1+\frac{f(x)}{x^2}\right)^x \left(1+\frac{2}{x}+o!\left(\frac{1}{x}\right)\right)^x.\]

Using the standard limit
\[\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x = e^{a},\]
we obtain
\[\lim_{x\to\infty}\left(1+\frac{f(x)}{x^2}\right)^x e^2.\]

Final answer

\[\boxed{e^2 \approx 7.389}\]