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# 数学代写|实分析代写Real Analysis代考|Indeterminate forms

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## 数学代写|实分析代写Real Analysis代考|Indeterminate forms

In the chapter on limits it was shown that if $\lim {x \rightarrow c} f(x)=l$ and $\lim {x \rightarrow c} g(x)=m \neq 0$ then $\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\frac{l}{m}$.

If however, $m=0$ then the limit could not be evaluated. The case when $l=0$ and $m=0$ was not covered in earlier chapters. In this case the limit of the quotient $\frac{f}{g}$ is said to take the indeterminate form $\frac{0}{0}$
We will see that in this case the limit may be finite or infinite, or even the limit may not exist.

The other indeterminate forms are represented by the symbols $\frac{\infty}{\infty}, \infty-\infty, 0 . \infty, 0^0, 1^{\infty}, 1^{-\infty}, \infty^0$.

We now discuss several theorems concerning evaluation of indeterminate forms.
Theorem 9.9.1. Case $\frac{0}{0}$
Let $c \in \mathbb{R}$. Let $f$ and $g$ be two functions such that $f(c)=g(c)=$ $0, g(x) \neq 0$ in some deleted neighbourhood $N^{\prime}(c, \delta) ; f$ and $g$ are differen tiable at $c$ and $g^{\prime}(c) \neq 0$. Then $\lim {x \rightarrow c} \frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}$. Proof. Let $x \in(c, c+\delta)$. Then $\frac{f(x)}{g(x)}=\frac{\frac{f(x)-f(c)}{f-c}}{\frac{g(x)-g(c)}{x-c}}$. Since $f$ and $g$ are differentiable at $c, \lim {x \rightarrow c+} \frac{f(x)-f(c)}{x-c}=R f^{\prime}(c)=$ $f^{\prime}(c)$ and $\lim {x \rightarrow c+} \frac{g(x)-g(c)}{x-c}=R g^{\prime}(c)=g^{\prime}(c)$. Therefore $\lim {x \rightarrow c+} \frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}$ since $g^{\prime}(c) \neq 0 \ldots \ldots$
Let $x \in(c-\delta, c)$.
Since $f$ and $g$ are differentiable at $c, \lim {x \rightarrow c-} \frac{f(x)-f(c)}{x-c}=L f^{\prime}(c)=$ $f^{\prime}(c)$ and $\lim {x \rightarrow c-} \frac{g(x)-g(c)}{x-c}=L g^{\prime}(c)=g^{\prime}(c)$.
Therefore $\lim {x \rightarrow c-} \frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}$ since $g^{\prime}(c) \neq 0 \ldots \ldots$ From (i) and (ii) we have $\lim {x \rightarrow c} \frac{f(x)}{g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}$.

## 数学代写|实分析代写Real Analysis代考|Worked Examples

Evaluate $\lim {x \rightarrow 0+}\left(\frac{1}{x}-\frac{1}{\sin x}\right)$. Let $f(x)=\frac{1}{x}, x \in(0,1), g(x)=\frac{1}{\sin x}, x \in(0,1)$. $\lim {x \rightarrow 0+}[f(x)-g(x)]$ takes the indeterminate form $\infty-\infty$.
We have $\lim {x \rightarrow 0+}\left(\frac{1}{x}-\frac{1}{\sin x}\right)=\lim {x \rightarrow 0+} \frac{\sin x-x}{x \sin x} \quad\left(=\frac{0}{0}\right)$
\begin{aligned} & =\lim {x \rightarrow 0+} \frac{\cos x-1}{\sin x+x \cos x} \quad\left(=\frac{0}{0}\right) \ & =\lim {x \rightarrow 0+} \frac{-\sin x}{2 \cos x-x \sin x}=0 . \end{aligned}

Evaluate $\lim {x \rightarrow 0+} x \log x$. Let $f(x)=x, x \in(0, \infty), g(x)=\log x, x \in(0, \infty)$. Then $\lim {x \rightarrow 0^{+}} x=0, \lim {x \rightarrow 0^{+}} \log x=\infty$. $\lim {x \rightarrow 0^{+}} x \log x$ takes the indeterminate form $0 . \infty$.
We have $\lim {x \rightarrow 0+} x \log x=\lim {x \rightarrow 0^{+}} \frac{\log x}{\frac{1}{x}} \quad\left(=\frac{\infty}{\infty}\right)$
$$=\lim {x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim {x \rightarrow 0^{+}}(-x)=0 .$$

Evaluate $\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$.
Let $f(x)=\frac{\sin x}{x}, x \neq 0, g(x)=\frac{1}{x}, x \neq 0$.

Then $\lim {x \rightarrow 0+} f(x)=1, \lim {x \rightarrow 0+} g(x)=\infty$.
$\lim {x \rightarrow 0+} f(x)^{g(x)}$ takes the indeterminate form $1^{\infty}$. \begin{aligned} & \lim {x \rightarrow 0+} \log \left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=\lim {x \rightarrow 0+} \frac{\log \frac{\sin x}{x}}{x} \quad\left(=\frac{0}{0}\right) \ &=\lim {x \rightarrow 0+} \frac{\frac{x}{\sin x} \cdot \frac{x \cos x-\sin x}{x^2}}{1} \ &=\lim {x \rightarrow 0+} \frac{x \cos x-\sin x}{x \sin x} \quad\left(=\frac{0}{0}\right) \ &=\lim {x \rightarrow 0+} \frac{-x \sin x}{x \cos x+\sin x}\left(=\frac{0}{0}\right) \ &=\lim {x \rightarrow 0+} \frac{-x \cos x-\sin x}{-x \sin x+2 \cos x}=0 . \end{aligned} Therefore $\lim {x \rightarrow 0+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^0=1$.
Also we have $\lim {x \rightarrow 0^{-}} f(x)=1$ and $\lim {x \rightarrow 0-} g(x)=-\infty$. $\lim {x \rightarrow 0-} f(x)^{g(x)}$ takes the indeterminate form $1^{-\infty}$. Proceeding similarly, we have $\lim {x \rightarrow 0-}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=1$.
Consequently, $\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=1$.

Evaluate $\lim {x \rightarrow 0^{+}} x^x$. Let $f(x)=x, x>0 ; g(x)=x, x>0$. Then $\lim {x \rightarrow 0^{+}} f(x)=$ $0, \lim {x \rightarrow 0^{+}} g(x)=0$. $\lim {x \rightarrow 0+}[f(x)]^{g(x)}$ takes the indeterminate form $0^0$.
$$\lim {x \rightarrow 0+} \log x^x=\lim {x \rightarrow 0+} \frac{\log x}{\frac{1}{x}} \quad\left(=\frac{\infty}{\infty}\right)=\lim {x \rightarrow 0+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim {x \rightarrow 0+}-x=0 .$$
Therefore $\lim _{x \rightarrow 0+} x^x=1^x$.

## 数学代写|实分析代写Real Analysis代考|Indeterminate forms

$$\lim {x \rightarrow 0+} \log x^x=\lim {x \rightarrow 0+} \frac{\log x}{\frac{1}{x}} \quad\left(=\frac{\infty}{\infty}\right)=\lim {x \rightarrow 0+} \frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim {x \rightarrow 0+}-x=0 .$$

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