some typos.
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@@ -115,7 +115,7 @@ $$
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\lim_{x \rightarrow 0} \frac{e^x - e^{-x}}{x}.
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$$
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It too is of the indeterminate form $0/0$. The derivative of the top is $e^x + e^{-x}$, which is $2$ when $x=0$, so the ratio of $f'(0)/g'(0)$ is seen to be $2$ By continuity, the limit of the ratio of the derivatives is $2$. Then by L'Hospital's rule, the limit above is $2$.
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It too is of the indeterminate form $0/0$. The derivative of the top is $e^x + e^{-x}$, which is $2$ when $x=0$, so the ratio of $f'(0)/g'(0)$ is seen to be $2$. By continuity, the limit of the ratio of the derivatives is $2$. Then by L'Hospital's rule, the limit above is $2$.
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* Sometimes, L'Hospital's rule must be applied twice. Consider this limit:
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@@ -820,7 +820,7 @@ $$
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#| hold: true
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#| echo: false
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choices = [
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"``e^{-2/\\pi}``",
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"``e^{2/\\pi}``",
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"``{2\\pi}``",
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"``1``",
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"``0``",
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