PUNE MCA Exam 2015 Question Paper with answers for online practice

Concept:Given $\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1$, only differences derived from continuous transformations preserving asymptotic equivalence are guaranteed to have limit $0$ under additional conditions, but here none are guaranteed except when the transformation is uniformly continuous on the range. The exponential difference does not follow because $e^f/e^g = e^{f-g}$ need not tend to $1$.Explanation:Since $\frac{f}{g} \to 1$, we have $f \sim g$. For any power or root, $f^a \sim g^a$, so $\frac{f^2}{g^2} \to 1$ and $\frac{\sqrt{f}}{\sqrt{g}} \to 1$. But $\frac{e^f}{e^g} = e^{f-g}$ does not necessarily tend to $1$ because $f-g$ may not tend to $0$ (e.g., $f=x+1$, $g=x$ gives $f-g=1$). Thus $e^f - e^g$ is not a consequence of $f/g\to1$, while $f^2-g^2$ and $\sqrt{f}-\sqrt{g}$ might approach $0$ in some cases but not always; however, the question asks for the option that is not a consequence, and $e^f-e^g$ is the most clearly non‑guaranteed when considering the ratio.Option D simplifies to $-(f+g)$, which is obviously unrelated.Given the typical exam context, the intended answer is the exponential difference.Answer:Option C: $e^f - e^g$ is not a consequence of $\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1$.