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-    "markdown": "# L'Hospital's Rule\n\n\n\nThis section uses these add-on packages:\n\n``` {.julia .cell-code}\nusing CalculusWithJulia\nusing Plots\nusing SymPy\n```\n\n\n\n\n---\n\n\nLet's return to limits of the form $\\lim_{x \\rightarrow c}f(x)/g(x)$ which have an indeterminate form of $0/0$ if both are evaluated at $c$. The typical example being the limit considered by Euler:\n\n\n\n$$\n\\lim_{x\\rightarrow 0} \\frac{\\sin(x)}{x}.\n$$\n\n\nWe know this is $1$ using a bound from geometry, but might also guess this is one, as we know from linearization near $0$ that we have $\\sin(x) \\approx x$ or, more specifically:\n\n\n\n$$\n\\sin(x) = x - \\sin(\\xi)x^2/2, \\quad 0 < \\xi < x.\n$$\n\n\nThis would yield:\n\n\n\n$$\n\\lim_{x \\rightarrow 0} \\frac{\\sin(x)}{x} = \\lim_{x\\rightarrow 0} \\frac{x -\\sin(\\xi) x^2/2}{x} = \\lim_{x\\rightarrow 0} 1 + \\sin(\\xi) \\cdot x/2 = 1.\n$$\n\n\nThis is because we know $\\sin(\\xi) x/2$  has a limit of $0$, when $|\\xi| \\leq |x|$.\n\n\nThat doesn't look any easier, as we worried about the error term, but if just mentally replaced $\\sin(x)$ with $x$ - which it basically is near $0$ - then we can see that the limit should be the same as $x/x$ which we know is $1$ without thinking.\n\n\nBasically, we found that in terms of limits, if both $f(x)$ and $g(x)$ are $0$ at $c$, that we *might* be able to just take this limit: $(f(c) + f'(c) \\cdot(x-c)) / (g(c) + g'(c) \\cdot (x-c))$ which is just $f'(c)/g'(c)$.\n\n\nWouldn't that be nice? We could find difficult limits just by differentiating the top and the bottom at $c$ (and not use the messy quotient rule).\n\n\nWell, in fact that is more or less true, a fact that dates back to [L'Hospital](http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule) - who wrote the first textbook on differential calculus - though this result is likely due to one of the Bernoulli brothers.\n\n\n> *L'Hospital's rule*: Suppose:\n>\n>   * that $\\lim_{x\\rightarrow c+} f(c) =0$ and $\\lim_{x\\rightarrow c+} g(c) =0$,\n>   * that $f$ and $g$ are differentiable in $(c,b)$, and\n>   * that $g(x)$ exists and is non-zero for *all* $x$ in $(c,b)$,\n>\n> then **if** the following limit exists: $\\lim_{x\\rightarrow c+}f'(x)/g'(x)=L$ it follows that $\\lim_{x \\rightarrow c+}f(x)/g(x) = L$.\n\n\n\nThat is *if* the right limit of $f(x)/g(x)$ is indeterminate of the form $0/0$, but the right limit of $f'(x)/g'(x)$ is known, possibly by simple continuity, then the right limit of $f(x)/g(x)$ exists and is equal to that of $f'(x)/g'(x)$.\n\n\nThe rule equally applies to *left limits* and *limits* at $c$. Later it will see there are other generalizations.\n\n\nTo apply this rule to Euler's example, $\\sin(x)/x$, we just need to consider that:\n\n\n\n$$\nL = 1 = \\lim_{x \\rightarrow 0}\\frac{\\cos(x)}{1},\n$$\n\n\nSo, as well, $\\lim_{x \\rightarrow 0} \\sin(x)/x = 1$.\n\n\nThis is due to $\\cos(x)$ being continuous at $0$, so this limit is just $\\cos(0)/1$. (More importantly, the tangent line expansion of $\\sin(x)$ at $0$ is $\\sin(0) + \\cos(0)x$, so that $\\cos(0)$ is why this answer is as it is, but we don't need to think in terms of $\\cos(0)$, but rather the tangent-line expansion, which is $\\sin(x) \\approx x$, as $\\cos(0)$ appears as the coefficient.\n\n\n:::{.callout-note}\n## Note\nIn [Gruntz](http://www.cybertester.com/data/gruntz.pdf), in a reference attributed to Speiss, we learn that L'Hospital was a French Marquis who was taught in $1692$ the calculus of Leibniz by Johann Bernoulli. They made a contract obliging Bernoulli to leave his mathematical inventions to L'Hospital in exchange for a regular compensation. This result was discovered in $1694$ and appeared in L'Hospital's book of $1696$.\n\n:::\n\n##### Examples\n\n\n  * Consider this limit at $0$: $(a^x - 1)/x$. We have $f(x) =a^x-1$ has $f(0) = 0$, so this limit is indeterminate of the form $0/0$. The derivative of $f(x)$ is $f'(x) = a^x \\log(a)$ which has $f'(0) = \\log(a)$. The derivative of the bottom is also $1$ at $0$, so we have:\n\n\n\n$$\n\\log(a) = \\frac{\\log(a)}{1} = \\frac{f'(0)}{g'(0)} = \\lim_{x \\rightarrow 0}\\frac{f'(x)}{g'(x)} = \\lim_{x \\rightarrow 0}\\frac{f(x)}{g(x)}\n= \\lim_{x \\rightarrow 0}\\frac{a^x - 1}{x}.\n$$\n\n\n:::{.callout-note}\n## Note\nWhy rewrite in the \"opposite\" direction? Because the theorem's result – $L$ is the limit – is only true if the related limit involving the derivative exists. We don't do this in the following, but did so here to emphasize the need for the limit of the ratio of the derivatives to exist.\n\n:::\n\n  * Consider this limit:\n\n\n\n$$\n\\lim_{x \\rightarrow 0} \\frac{e^x - e^{-x}}{x}.\n$$\n\n\nIt 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 $2$.\n\n\n  * Sometimes, L'Hospital's rule mustLine truncated
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