Update Math4121_L4.md
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@@ -90,12 +90,14 @@ Case 1: $f(x)\to 0$ and $g(x)\to 0$ as $x\to a$.
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As $x\to a$, $f(x)\to 0$ and $g(x)\to 0$. So
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As $x\to a$, $f(x)\to 0$ and $g(x)\to 0$. So
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$$\begin{aligned}
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$$
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\begin{aligned}
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\lim_{x\to a}\frac{f(x)-f(y)}{g(x)-g(y)}&=\lim_{x\to a}\frac{0-f(y)}{0-g(y)}\\
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\lim_{x\to a}\frac{f(x)-f(y)}{g(x)-g(y)}&=\lim_{x\to a}\frac{0-f(y)}{0-g(y)}\\
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&=\lim_{x\to a}\frac{f(y)}{g(y)}\\
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&=\lim_{x\to a}\frac{f(y)}{g(y)}\\
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&=\frac{f'(y)}{g'(y)}\\
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&=\frac{f'(y)}{g'(y)}\\
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&\leq r<q
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&\leq r<q
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\end{aligned}$$
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\end{aligned}
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$$
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$\forall y\in (a,c)$, $\frac{f(y)}{g(y)}<q$.
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$\forall y\in (a,c)$, $\frac{f(y)}{g(y)}<q$.
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