Question

Let f : (−1, 1) → \(\R\) and g : (−1, 1) → \(\R\) be thrice continuously differentiable functions such that f(x) ≠ g(x) for every nonzero x ∈ (−1, 1). Suppose
f(0) = ln 2, f'(0) = π, f"(0) = π² , and f"'(0) = π⁹
and
g(0) = ln 2, g'(0) = π, g′′(0) = π², and g'"(0) = π³.
Then the value of the limit
\(\lim\limits_{x\rightarrow0}\frac{e^{f(x)}-e^{g(x)}}{f(x)-g(x)}\)
is equal to _________. (Rounded off to two decimal places)

Answer 1

Correct Answer: 2

We need to evaluate the limit: \(\lim\limits_{x\rightarrow0}\frac{e^{f(x)}-e^{g(x)}}{f(x)-g(x)}\). This resembles the form where L'Hôpital's Rule can be applied if direct substitution results in an indeterminate form. Start by noting:
\(e^{f(x)}\approx e^{\ln 2 + \pi x + \frac{\pi^2}{2} x^2 + \frac{\pi^9}{6} x^3}\) and 
\(e^{g(x)}\approx e^{\ln 2 + \pi x + \frac{\pi^2}{2} x^2 + \frac{\pi^3}{6} x^3}\).
Using series expansions:
\(e^{f(x)} - e^{g(x)} \approx \left(e^{f(0)} + e^{f(0)}(\pi x + \frac{\pi^2}{2} x^2 + \frac{\pi^9}{6} x^3) - (e^{g(0)} + e^{g(0)}(\pi x + \frac{\pi^2}{2} x^2 + \frac{\pi^3}{6} x^3))\right)\).
This reduces to:
\((e^{\ln 2}) \cdot \frac{(\pi^9-\pi^3)}{6} x^3\).
The denominator is:
\(f(x)-g(x) \approx \left(\ln2 + \pi x + \frac{\pi^2}{2} x^2 + \frac{\pi^9}{6} x^3\right) - \left(\ln2 + \pi x + \frac{\pi^2}{2} x^2 + \frac{\pi^3}{6} x^3\right) = \frac{\pi^9-\pi^3}{6}x^3\).
Substitute these into the limit:
\(\lim\limits_{x\rightarrow0}\frac{(e^{\ln 2}) \cdot \frac{(\pi^9-\pi^3)}{6} x^3}{\frac{\pi^9-\pi^3}{6} x^3} = e^{\ln 2} = 2\).
The computed limit is 2.00.