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BODY
{
FONT-FAMILY: NYT Imperial + Noto Sans SC;
FONT-SIZE: 18pt;
LINE-HEIGHT : 2.0em;
overflow: scroll;
height: 100%;
}
html {
height: 100%;
width: 100%;
overflow: hidden;
}
table {
margin: 0.5rem 0.5rem 0.5rem 0rem;
border-collapse: collapse;
font-size: 14pt;
FONT-FAMILY: Times Ten LT Std Roman;
FONT-SIZE: 16pt;
LINE-HEIGHT : 2.0em;
}
table th {
background:#5A97D5;
color: white;
width: auto;
FONT-FAMILY: Times Ten LT Std Roman;
FONT-SIZE: 16pt;
LINE-HEIGHT : 2.0em;
}
td {
width: auto;
border: 1px solid;
border-color:black;
padding:0 0.5em;
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BODY1
{
COLOR: white;
BACKGROUND-COLOR: #000080;
}
a:link {color:#000099}
a:visited {color:#000066}
a:hover {color:#6600CC}
a:active {color:#3333FF}
ul {
list-style-type:decimal;
border-left:2pt solid #e5e5e4;
position:relative;
padding-left:35pt;
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margin-top:5pt;
margin-bottom:5pt;
margin-left:2.5pt;
LINE-HEIGHT : 2.5em;
margin:10px;
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ol {
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position:relative;
padding-left:35pt;
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float:right;
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left: 5px;
top: 5px;
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color:#F0F0F0;
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top:-9999px;
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float:right;
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margin-right:.3em;
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COLOR:#000000;
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.schedule {
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}
$$
\begin{aligned}
\lim _{x \rightarrow 0} f^{\prime}(x) &=\lim _{x \rightarrow 0} \frac{\left[g^{\prime}(x)+\mathrm{e}^{-x}\right] x-\left[g(x)-\mathrm{e}^{-x}\right]}{x^{2}} \\
&=\lim _{x \rightarrow 0} \frac{g^{\prime}(x)+\mathrm{e}^{-x}}{x}-\lim _{x \rightarrow 0} \frac{g(x)-\mathrm{e}^{-x}}{x^{2}} \\
& \stackrel{L^{\prime}}{ } \lim _{x \rightarrow 0}\left[g^{\prime \prime}(x)-\mathrm{e}^{-x}\right]-\lim _{x \rightarrow 0} \frac{g^{\prime}(x)+\mathrm{e}^{-x}}{2 x} \\
&=g^{\prime \prime}(0)-1-\frac{1}{2}\left[g^{\prime \prime}(0)-1\right] \\
&=\frac{1}{2}\left[g^{\prime \prime}(0)-1\right]=f^{\prime}(0),
\end{aligned}
$$
故 $f^{\prime}(x)$ 在 $(-\infty,+\infty)$ 上连续.
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