Limit Puzzle

Calculus Level 3

Two functions $f(x)$ and $g(x)$ exist such that $f'(x), f''(x), g'(x), g''(x)$ are all equal to $0$ at $x = k.$

Given $f'(x) g(x)^{2} = \big(f(x) - g(x)\big)^{2} g'(x)\quad \text{ and }\quad f''(x) g(x)^{2} = \left(f(x) - \big(2 + \sqrt{5}\big) g(x)\right)^{2} g''(x),$ find the value of the following expression to 3 decimal places: $\lim_{x \to k} \frac{f(x)}{g(x)}.$

The answer is 2.618.

1 solution

I F
May 7, 2018

Rearranging the first equation gives:

$\frac{f'(x)}{g'(x)} = (\frac{f(x)}{g(x)} - 1)^{2}$

Taking limits on both sides:

$\lim_{x \to k} \frac{f'(x)}{g'(x)} = \lim_{x \to k}(\frac{f(x)}{g(x)} - 1)^{2}$

Since limits are commutable with continuous functions, we can rearrange to:

$\lim_{x \to k} \frac{f'(x)}{g'(x)} = (\lim_{x \to k}(\frac{f(x)}{g(x)}) - 1)^{2}$

The second equation can be similarly rearranged to:

$\lim_{x \to k} \frac{f''(x)}{g''(x)} = (\lim_{x \to k}(\frac{f(x)}{g(x)}) - (2 + \sqrt{5}))^{2}$

Since $f'(k)$ , $f''(k)$ , $g'(k)$ and $g''(k)$ are all zero, we can say by L'Hopital's rule that:

$\lim_{x \to k} \frac{f'(x)}{g'(x)} = \lim_{x \to k} \frac{f''(x)}{g''(x)}$

And therefore:

$(\lim_{x \to k}(\frac{f(x)}{g(x)}) - 1)^{2} = (\lim_{x \to k}(\frac{f(x)}{g(x)}) - (2 + \sqrt{5}))^{2}$

Solving this equation gives:

$\lim_{x \to k}(\frac{f(x)}{g(x)}) = (3 + \sqrt{5})/2 \approx \boxed{2.618}$

Limit Puzzle

The answer is 2.618.

1 solution

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