如果你也在 怎样代写微积分Calculus 这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。微积分Calculus 最初被称为无穷小微积分或 “无穷小的微积分”，是对连续变化的数学研究，就像几何学是对形状的研究，而代数是对算术运算的概括研究一样。

微积分Calculus 它有两个主要分支，微分和积分；微分涉及瞬时变化率和曲线的斜率，而积分涉及数量的累积，以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互关联，它们利用了无限序列和无限数列收敛到一个明确定义的极限的基本概念 。17世纪末，牛顿（Isaac Newton）和莱布尼兹（Gottfried Wilhelm Leibniz）独立开发了无限小数微积分。后来的工作，包括对极限概念的编纂，将这些发展置于更坚实的概念基础上。今天，微积分在科学、工程和社会科学中得到了广泛的应用。

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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 $0 / 0$

如果我们想知道这个函数
$$
F(x)=\frac{3 x-\sin x}{x}
$$
在$x=0$(未定义)附近的行为，我们可以检查$F(x)$的极限为$x \rightarrow 0$。由于分母的极限为0，所以不能用除法定则求极限(第二章定理1)。而且，在这种情况下，分子和分母都趋近于0,$0 / 0$没有定义。一般情况下，这样的极限可能存在，也可能不存在，但是通过应用’Hôpital规则，讨论的函数$F(x)$确实存在极限，正如我们将在示例1d中看到的那样。
如果连续函数$f(x)$和$g(x)$在$x=a$处均为零，则
通过替换$x=a$无法找到$$
\lim {x \rightarrow a} \frac{f(x)}{g(x)} $$。替换产生$0 / 0$，这是一个没有意义的表达式，我们无法求值。我们使用$0 / 0$作为没有数值的表达式的表示法，称为不定式。其他无意义的表达式也经常出现，例如$\infty / \infty, \infty \cdot 0, \infty-\infty, 0^0$和$1^{\infty}$，它们不能以一致的方式求值;这些也被称为不定式。有时，但并非总是如此，可以通过消去、重新排列项或其他代数操作来找到导致不定式形式的限制。这是我们在第二章中的经验。在2.4节中，我们进行了大量的分析才找到$\lim {x \rightarrow 0}(\sin x) / x$。但我们在限制方面取得了成功
$$
f^{\prime}(a)=\lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a},
$$
我们用它来计算导数如果我们试图代入$x=a$，它会产生不定式$0 / 0$。L’Hôpital规则使我们能够利用导数的成功经验来评估代换导致不定式的极限。

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

在证明l’Hôpital规则之前，我们考虑一个特殊情况，以对其合理性提供一些几何见解。考虑两个函数$f(x)$和$g(x)$具有连续导数并且满足$f(a)=g(a)=0, g^{\prime}(a) \neq 0$。$f(x)$和$g(x)$的图形以及它们的线性化$y=f^{\prime}(a)(x-a)$和$y=g^{\prime}(a)(x-a)$如图7.19所示。我们知道在$x=a$附近，线性化提供了很好的近似函数。事实上，
$$
f(x)=f^{\prime}(a)(x-a)+\varepsilon_1(x-a) \text { and } g(x)=g^{\prime}(a)(x-a)+\varepsilon_2(x-a)
$$
其中$\varepsilon_1 \rightarrow 0$和$\varepsilon_2 \rightarrow 0$为$x \rightarrow a$。因此，如图7.19所示，
$$
\begin{array}{rlr}
\lim {x \rightarrow a} \frac{f(x)}{g(x)} & =\lim {x \rightarrow a} \frac{f^{\prime}(a)(x-a)+\varepsilon_1(x-a)}{g^{\prime}(a)(x-a)+\varepsilon_2(x-a)} & \
& =\lim {x \rightarrow a} \frac{f^{\prime}(a)+\varepsilon_1}{g^{\prime}(a)+\varepsilon_2}=\frac{f^{\prime}(a)}{g^{\prime}(a)} & g^{\prime}(a) \neq 0 \ & =\lim {x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, & \text { Continuous derivatives }
\end{array}
$$
正如’Hôpital法则所断言的那样。现在我们根据定理5中更一般的假设来证明这个规则，它不需要$g^{\prime}(a) \neq 0$，也不需要两个函数有连续的导数。

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。
定理6:柯西中值定理
假设函数$f$和$g$在$[a, b]$上是连续的在$(a, b)$上是可微的也假设$g^{\prime}(x) \neq 0$在$(a, b)$上是可微的。然后在$(a, b)$中存在一个数字$c$
$$
\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}
$$

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