# 数学代写|微积分代写Calculus代写|Indeterminate Form $0 / 0$

#### Doug I. Jones

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## 数学代写|微积分代写Calculus代写|Indeterminate Form $0 / 0$

If we want to know how the function
$$F(x)=\frac{3 x-\sin x}{x}$$
behaves near $x=0$ (where it is undefined), we can examine the limit of $F(x)$ as $x \rightarrow 0$. We cannot apply the Quotient Rule for limits (Theorem 1 of Chapter 2) because the limit of the denominator is 0 . Moreover, in this case, both the numerator and denominator approach 0 , and $0 / 0$ is undefined. Such limits may or may not exist in general, but the limit does exist for the function $F(x)$ under discussion by applying l’Hôpital’s Rule, as we will see in Example 1d.
If the continuous functions $f(x)$ and $g(x)$ are both zero at $x=a$, then
$$\lim {x \rightarrow a} \frac{f(x)}{g(x)}$$ cannot be found by substituting $x=a$. The substitution produces $0 / 0$, a meaningless expression, which we cannot evaluate. We use $0 / 0$ as a notation for an expression that does not have a numerical value, known as an indeterminate form. Other meaningless expressions often occur, such as $\infty / \infty, \infty \cdot 0, \infty-\infty, 0^0$, and $1^{\infty}$, which cannot be evaluated in a consistent way; these are called indeterminate forms as well. Sometimes, but not always, limits that lead to indeterminate forms may be found by cancelation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find $\lim {x \rightarrow 0}(\sin x) / x$. But we have had success with the limit
$$f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a},$$
from which we calculate derivatives and which produces the indeterminant form $0 / 0$ if we attempt to substitute $x=a$. L’Hôpital’s Rule enables us to draw on our success with derivatives to evaluate limits for which substitution leads to indeterminate forms.

## 数学代写|微积分代写Calculus代写|Proof of L’Hôpital’s Rule

Before we prove l’Hôpital’s Rule, we consider a special case to provide some geometric insight for its reasonableness. Consider the two functions $f(x)$ and $g(x)$ having continuous derivatives and satisfying $f(a)=g(a)=0, g^{\prime}(a) \neq 0$. The graphs of $f(x)$ and $g(x)$, together with their linearizations $y=f^{\prime}(a)(x-a)$ and $y=g^{\prime}(a)(x-a)$, are shown in Figure 7.19. We know that near $x=a$, the linearizations provide good approximations to the functions. In fact,
$$f(x)=f^{\prime}(a)(x-a)+\varepsilon_1(x-a) \text { and } g(x)=g^{\prime}(a)(x-a)+\varepsilon_2(x-a)$$
where $\varepsilon_1 \rightarrow 0$ and $\varepsilon_2 \rightarrow 0$ as $x \rightarrow a$. So, as Figure 7.19 suggests,
$$\begin{array}{rlr} \lim {x \rightarrow a} \frac{f(x)}{g(x)} & =\lim {x \rightarrow a} \frac{f^{\prime}(a)(x-a)+\varepsilon_1(x-a)}{g^{\prime}(a)(x-a)+\varepsilon_2(x-a)} & \ & =\lim {x \rightarrow a} \frac{f^{\prime}(a)+\varepsilon_1}{g^{\prime}(a)+\varepsilon_2}=\frac{f^{\prime}(a)}{g^{\prime}(a)} & g^{\prime}(a) \neq 0 \ & =\lim {x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, & \text { Continuous derivatives } \end{array}$$
as asserted by l’Hôpital’s Rule. We now proceed to a proof of the rule based on the more general assumptions stated in Theorem 5, which do not require that $g^{\prime}(a) \neq 0$ and that the two functions have continuous derivatives.

The proof of l’Hôpital’s Rule is based on Cauchy’s Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Cauchy’s Theorem first and then show how it leads to l’Hôpital’s Rule.
THEOREM 6-Cauchy’s Mean Value Theorem
Suppose functions $f$ and $g$ are continuous on $[a, b]$ and differentiable throughout $(a, b)$ and also suppose $g^{\prime}(x) \neq 0$ throughout $(a, b)$. Then there exists a number $c$ in $(a, b)$ at which
$$\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Indeterminate Form $0 / 0$

$$F(x)=\frac{3 x-\sin x}{x}$$

$$f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a},$$

## 数学代写|微积分代写Calculus代写|Proof of L’Hôpital’s Rule

$$f(x)=f^{\prime}(a)(x-a)+\varepsilon_1(x-a) \text { and } g(x)=g^{\prime}(a)(x-a)+\varepsilon_2(x-a)$$

$$\begin{array}{rlr} \lim {x \rightarrow a} \frac{f(x)}{g(x)} & =\lim {x \rightarrow a} \frac{f^{\prime}(a)(x-a)+\varepsilon_1(x-a)}{g^{\prime}(a)(x-a)+\varepsilon_2(x-a)} & \ & =\lim {x \rightarrow a} \frac{f^{\prime}(a)+\varepsilon_1}{g^{\prime}(a)+\varepsilon_2}=\frac{f^{\prime}(a)}{g^{\prime}(a)} & g^{\prime}(a) \neq 0 \ & =\lim {x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, & \text { Continuous derivatives } \end{array}$$

l’Hôpital规则的证明基于柯西中值定理，这是中值定理的扩展，涉及两个函数而不是一个函数。我们首先证明柯西定理，然后说明它是如何引出’Hôpital法则的。

$$\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$$

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