Question

L'Hôpital's Rule by Taylor series Suppose f and g have Taylor series about the point a.
a. If f(a) = g(a) = 0 and g′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent withl’Hôpital’s Rule.
b. If f(a) = g(a) =f′(a) = g′(a) = 0 and g′′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent with two applications of 1'Hôpital's Rule.

Accepted Answer

Step 1: Recall the Taylor series expansions of functions $$f$$ and $$g$$ about the point $$a. $$For a function $$h$$, the expansion is:

$$
\displaystyle h(x) = h(a) + h'(a)(x - a) + \frac{h''(a)}{2!}(x - a)^2 + \cdots

We $$will use this to expand both $$f(x)$$ and $$g(x). $$Step 2: For part (a), since $$f(a) = g(a) = 0$$ and $$g'(a) 
eq 0$$, write the expansions:

$$
\begin{aligned}
f(x) &= f(a) + f'(a)(x - a) + \cdots = f'(a)(x - a) + \cdots \
g(x) &= g(a) + g'(a)(x - a) + \cdots = g'(a)(x - a) + \cdots
\end{aligned}
$$

Ignore higher order terms as $$x 	o a. $$Step 3: Form the ratio $$\frac{f(x)}{g(x)}$$ near $$x = a$$:

$$
\frac{f(x)}{g(x)} \approx \frac{f'(a)(x - a)}{g'(a)(x - a)} = \frac{f'(a)}{g'(a)}
$$

This shows the limit is $$\lim_{x 	o a} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}$$, consistent with L'Hôpital's Rule applied once. Step 4: For part (b), given $$f(a) = g(a) = f'(a) = g'(a) = 0$$ and $$g''(a) 
eq 0$$, write the expansions including second derivatives:

$$
\begin{aligned}
f(x) &= f(a) + f'(a)(x - a) + \frac{f''(a)}{2}(x - a)^2 + \cdots = \frac{f''(a)}{2}(x - a)^2 + \cdots \
g(x) &= g(a) + g'(a)(x - a) + \frac{g''(a)}{2}(x - a)^2 + \cdots = \frac{g''(a)}{2}(x - a)^2 + \cdots
\end{aligned}
$$ Step 5: Form the ratio near $$x = a$$:

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

This matches the result of applying L'Hôpital's Rule twice, since the first derivatives vanish and the second derivatives determine the limit.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Taylor Series Expansion

L'Hôpital's Rule

Relationship Between Taylor Series and L'Hôpital's Rule